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MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Corrected exercises: systems of linear equations and Gaussian elimination (MATH 133)

This is the corrected exercise set for the first chapter of MATH 133, Linear Algebra and Geometry, at McGill University: systems of linear equations and Gaussian elimination, sections 1.1, 1.2 and 1.4 of Nicholson's Linear Algebra with Applications.

What the set covers, the traps it targets

It covers the augmented matrix, the three elementary row operations, the row echelon and reduced row echelon forms, pivot and free variables, the general solution in parametric form, inconsistent systems, systems with a parameter, and the classic applications: interpolation, circuits, chemical equations and mixtures. No calculator is needed anywhere, as on the exam: every pivot reduces by hand.

Exercise 0.1 opens the set with twelve systems to solve by Gaussian elimination, the plain list of a textbook: square systems with one solution, a first pivot equal to 00 that forces a row swap, systems with more equations than unknowns that are consistent or not, systems with one, two, three and four free variables written as a general solution, a system with more unknowns than equations and no solution, and equations with missing unknowns. Every answer is checked as you type it.

The thread running through the whole set: a row operation is legal only when it can be undone, and the answer is read off the echelon form and nowhere else. Every trap of the chapter breaks one of those two rules: dividing by k−2k - 2 before knowing that k≠2k \neq 2, replacing a row by a combination that no longer contains it, declaring infinitely many solutions because of a zero row, or calling a variable free because of where its column sits rather than because its column has no pivot.

The traps named explicitly in the solutions: a variable copied into the wrong column, a replaced row that no longer contains itself, two operations done at once on the old rows, a division by k−2k - 2 that silently removes the case k=2k = 2, a first step that divides by kk, a zero row taken as proof of infinitely many solutions, a free variable declared before the consistency test, a pivot variable used as a parameter, and a chemical equation handed in with fractional coefficients.

  • 11 corrected exercises • 110 points • 150 minutes
  • Self-checking, 155 answers
  • No calculator
Start exercise 0.1 The chapter sheet

Course recap
  • • Elementary row operations: swap two rows; multiply a row by a NON-ZERO constant; add a multiple of ANOTHER row. Each one can be undone, so the solution set never changes.
  • • Row echelon form: zero rows at the bottom, each leading entry strictly to the right of the one above. Reduced: in addition each leading entry is 11 and is the only non-zero entry of its column. The reduced form of a matrix is UNIQUE; an echelon form is not, but its pivot columns are.
  • • A system is inconsistent exactly when an echelon form has a row (0  ⋯  0∣c)(0 \; \cdots \; 0 \mid c) with c≠0c \neq 0.
  • • Pivot variables sit in columns with a leading entry, free variables in columns without one. A consistent system with a free variable has infinitely many solutions, one parameter per free column.
  • • With a parameter kk: let expressions in kk multiply the OTHER row only, never divide by one; factor the last pivot, then study each value where it vanishes separately, right-hand side included.
  • • Geometry: with two unknowns each equation is a line, with three a plane, and a solution is a point common to all of them.

Calculations: systems to solve by elimination

Exercise 0.1: Twelve systems to solve by Gaussian elimination

44 answers to check

For each system, write the augmented matrix, reduce it to row echelon form, then conclude: give the solution if it is unique, the general solution if there are infinitely many, or show that there is none. Some systems have more equations than unknowns, others more unknowns than equations: the shape alone never gives the answer. No calculator: every pivot reduces by hand.

Three row operations only: swap two rows, multiply a row by a non-zero number, add a multiple of another row. Clear the first column below the first pivot, then the next column below the next pivot, and read the result. For a system with free variables, the values asked for below are read from the general solution, the free variables being those whose column holds no pivot.

  • a) Solve the system. {x1−2x2−4x3=−162x1+2x2+2x3=14−2x1+3x2+3x3=21\begin{cases} x_{1} - 2x_{2} - 4x_{3} = -16 \\ 2x_{1} + 2x_{2} + 2x_{3} = 14 \\ -2x_{1} + 3x_{2} + 3x_{3} = 21 \end{cases}
  • b) Solve the system. {4x2+4x3=−44−x1−2x2+4x3=−11−2x1−3x2+4x3=−3\begin{cases} 4x_{2} + 4x_{3} = -44 \\ -x_{1} - 2x_{2} + 4x_{3} = -11 \\ -2x_{1} - 3x_{2} + 4x_{3} = -3 \end{cases}
  • c) How many solutions? {4x1+4x2+4x3=52−x1−2x2+4x3=17−2x1−3x2+4x3=11x1−2x2−3x3=−30\begin{cases} 4x_{1} + 4x_{2} + 4x_{3} = 52 \\ -x_{1} - 2x_{2} + 4x_{3} = 17 \\ -2x_{1} - 3x_{2} + 4x_{3} = 11 \\ x_{1} - 2x_{2} - 3x_{3} = -30 \end{cases}
  • d) How many solutions? {4x1+x2=222x1−2x2=−43x1−4x2=−122x1+4x2=32x1+4x2=28\begin{cases} 4x_{1} + x_{2} = 22 \\ 2x_{1} - 2x_{2} = -4 \\ 3x_{1} - 4x_{2} = -12 \\ 2x_{1} + 4x_{2} = 32 \\ x_{1} + 4x_{2} = 28 \end{cases}
  • e) How many solutions? {3x1−4x2−3x3=−20−x1+x2+2x3=92x1+3x2+x3=72x1−3x2−x3=−7\begin{cases} 3x_{1} - 4x_{2} - 3x_{3} = -20 \\ -x_{1} + x_{2} + 2x_{3} = 9 \\ 2x_{1} + 3x_{2} + x_{3} = 7 \\ 2x_{1} - 3x_{2} - x_{3} = -7 \end{cases}
  • f) Solve the system. {−4x1−3x2+x4=02x1+3x2−3x4=22−x1+2x3+2x4=−123x2+3x3−4x4=34\begin{cases} -4x_{1} - 3x_{2} + x_{4} = 0 \\ 2x_{1} + 3x_{2} - 3x_{4} = 22 \\ -x_{1} + 2x_{3} + 2x_{4} = -12 \\ 3x_{2} + 3x_{3} - 4x_{4} = 34 \end{cases}
  • g) Find the general solution. {−x1+x2+3x3−x4=27−3x1+2x2+3x3−2x4=45−x1+3x2−3x3−3x4=9\begin{cases} -x_{1} + x_{2} + 3x_{3} - x_{4} = 27 \\ -3x_{1} + 2x_{2} + 3x_{3} - 2x_{4} = 45 \\ -x_{1} + 3x_{2} - 3x_{3} - 3x_{4} = 9 \end{cases}
  • h) Find the general solution. {−4x1−2x2−4x3+4x4=12x1−x2−2x3−x4=63x1+x3−3x4−x5=−2\begin{cases} -4x_{1} - 2x_{2} - 4x_{3} + 4x_{4} = 12 \\ x_{1} - x_{2} - 2x_{3} - x_{4} = 6 \\ 3x_{1} + x_{3} - 3x_{4} - x_{5} = -2 \end{cases}
  • i) Find the general solution. {−2x1−x2+4x3−3x4+x5+2x6=8x1+x2−x3+x4−x5−x6=−2{\small \begin{cases} -2x_{1} - x_{2} + 4x_{3} - 3x_{4} + x_{5} + 2x_{6} = 8 \\ x_{1} + x_{2} - x_{3} + x_{4} - x_{5} - x_{6} = -2 \end{cases}}
  • j) Find the general solution. {x1−2x2−3x3−4x4−2x6=−6x1−x2−2x3−2x4+x5=−32x1−3x3+2x4−3x5+x6=−10{\small \begin{cases} x_{1} - 2x_{2} - 3x_{3} - 4x_{4} - 2x_{6} = -6 \\ x_{1} - x_{2} - 2x_{3} - 2x_{4} + x_{5} = -3 \\ 2x_{1} - 3x_{3} + 2x_{4} - 3x_{5} + x_{6} = -10 \end{cases}}
  • k) How many solutions? {3x1−2x2−x3−x4=0−x1−2x2−x3+2x4=8−2x1+4x2+2x3−x4=−12\begin{cases} 3x_{1} - 2x_{2} - x_{3} - x_{4} = 0 \\ -x_{1} - 2x_{2} - x_{3} + 2x_{4} = 8 \\ -2x_{1} + 4x_{2} + 2x_{3} - x_{4} = -12 \end{cases}
  • l) How many solutions? {4x1+4x2=−20−x1−2x2=12−2x1+4x3=20−2x1−3x2=173x1+4x2−2x3=−34\begin{cases} 4x_{1} + 4x_{2} = -20 \\ -x_{1} - 2x_{2} = 12 \\ -2x_{1} + 4x_{3} = 20 \\ -2x_{1} - 3x_{2} = 17 \\ 3x_{1} + 4x_{2} - 2x_{3} = -34 \end{cases}

Type your answers, the page tells you right or wrong 0/44

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Show the solution

Answers

  • a) one solution: (x1,x2,x3)=(0,6,1)(x_{1}, x_{2}, x_{3}) = (0, 6, 1)
  • b) one solution: (x1,x2,x3)=(−3,−5,−6)(x_{1}, x_{2}, x_{3}) = (-3, -5, -6)
  • c) one solution: (x1,x2,x3)=(1,5,7)(x_{1}, x_{2}, x_{3}) = (1, 5, 7)
  • d) one solution: (x1,x2)=(4,6)(x_{1}, x_{2}) = (4, 6)
  • e) row (0,0,0∣1)(0, 0, 0 \mid 1), so no solution
  • f) one solution: (x1,x2,x3,x4)=(−4,3,−1,−7)(x_{1}, x_{2}, x_{3}, x_{4}) = (-4, 3, -1, -7)
  • g) infinitely many solutions: (x1,x2,x3,x4)=(−6,6+t,5,t)(x_{1}, x_{2}, x_{3}, x_{4}) = (-6, 6 + t, 5, t), tt real
  • h) infinitely many solutions: (x1,x2,x3,x4,x5)=(s,−2−2t,−2+t,s,t)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (s, -2 - 2t, -2 + t, s, t), s,ts, t real
  • i) infinitely many solutions: (x1,x2,x3,x4,x5,x6)=(−6+3r−2s+u,4−2r+s+t,r,s,t,u)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}) = (-6 + 3r - 2s + u, 4 - 2r + s + t, r, s, t, u), r,s,t,ur, s, t, u real
  • j) infinitely many solutions: (x1,x2,x3,x4,x5,x6)=(10+2r−9s−5t,−7−4r+6s+t,10+2r−7s−3t,r,s,t)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}) = (10 + 2r - 9s - 5t, -7 - 4r + 6s + t, 10 + 2r - 7s - 3t, r, s, t), r,s,tr, s, t real
  • k) row (0,0,0,0∣−1)(0, 0, 0, 0 \mid -1), so no solution
  • l) one solution: (x1,x2,x3)=(2,−7,6)(x_{1}, x_{2}, x_{3}) = (2, -7, 6)

One gesture decides everything: row reduce, then read the LAST rows of the echelon form before concluding. A row (0,…,0∣c)(0, \dots, 0 \mid c) with c≠0c \neq 0 means no solution. Otherwise the system is consistent, and it has exactly one solution when every column of unknowns holds a pivot, infinitely many when at least one does not. The shape of the system (more equations, more unknowns) never decides on its own.

The operations below take a pivot 11 or −1-1 whenever one is available, bringing it up by a row swap, so that every entry stays an integer; dividing a row by a common factor keeps the numbers small. Another order of operations gives other intermediate matrices but the same conclusion.

a)

Augmented matrix:

(1−2−4−1622214−23321)\displaystyle \left(\begin{array}{ccc|c} 1 & -2 & -4 & -16 \\ 2 & 2 & 2 & 14 \\ -2 & 3 & 3 & 21 \end{array}\right)

Column 11: R2→R2−2R1R_{2} \to R_{2} - 2R_{1}, R3→R3+2R1R_{3} \to R_{3} + 2R_{1}, R2→12R2R_{2} \to \frac{1}{2}R_{2}.

(1−2−4−16035230−1−5−11)\displaystyle \left(\begin{array}{ccc|c} 1 & -2 & -4 & -16 \\ 0 & 3 & 5 & 23 \\ 0 & -1 & -5 & -11 \end{array}\right)

Column 22: R2↔R3R_{2} \leftrightarrow R_{3}, R3→R3+3R2R_{3} \to R_{3} + 3R_{2}, R3→110R3R_{3} \to \frac{1}{10}R_{3}.

(1−2−4−160−1−5−1100−1−1)\displaystyle \left(\begin{array}{ccc|c} 1 & -2 & -4 & -16 \\ 0 & -1 & -5 & -11 \\ 0 & 0 & -1 & -1 \end{array}\right)

Row 33: −x3=−1-x_{3} = -1, so x3=1x_{3} = 1.

Row 22: −x2−5x3=−11-x_{2} - 5x_{3} = -11, with x3=1x_{3} = 1, gives x2=6x_{2} = 6.

Row 11: x1−2x2−4x3=−16x_{1} - 2x_{2} - 4x_{3} = -16, with x2=6x_{2} = 6, x3=1x_{3} = 1, gives x1=0x_{1} = 0.

Check in the original system: 0−12−4=−160 - 12 - 4 = -16, 0+12+2=140 + 12 + 2 = 14, 0+18+3=210 + 18 + 3 = 21.

Pitfall: in R2→R2−2R1R_2 \to R_2 - 2R_1 the constant becomes 14−2(−16)=4614 - 2(-16) = 46, not 14−32=−1814 - 32 = -18: the minus sign of the operation meets the minus sign of the entry.

b)

Augmented matrix:

(044−44−1−24−11−2−34−3)\displaystyle \left(\begin{array}{ccc|c} 0 & 4 & 4 & -44 \\ -1 & -2 & 4 & -11 \\ -2 & -3 & 4 & -3 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R3→R3−2R1R_{3} \to R_{3} - 2R_{1}, R2→14R2R_{2} \to \frac{1}{4}R_{2}.

(−1−24−11011−1101−419)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 4 & -11 \\ 0 & 1 & 1 & -11 \\ 0 & 1 & -4 & 19 \end{array}\right)

Column 22: R3→R3−R2R_{3} \to R_{3} - R_{2}, R3→15R3R_{3} \to \frac{1}{5}R_{3}.

(−1−24−11011−1100−16)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 4 & -11 \\ 0 & 1 & 1 & -11 \\ 0 & 0 & -1 & 6 \end{array}\right)

Row 33: −x3=6-x_{3} = 6, so x3=−6x_{3} = -6.

Row 22: x2+x3=−11x_{2} + x_{3} = -11, with x3=−6x_{3} = -6, gives x2=−5x_{2} = -5.

Row 11: −x1−2x2+4x3=−11-x_{1} - 2x_{2} + 4x_{3} = -11, with x2=−5x_{2} = -5, x3=−6x_{3} = -6, gives x1=−3x_{1} = -3.

Check in the original system: −20−24=−44-20 - 24 = -44, 3+10−24=−113 + 10 - 24 = -11, 6+15−24=−36 + 15 - 24 = -3.

Pitfall: the first entry of R1R_1 is 00, so it cannot be a pivot. Swap two ROWS; swapping two columns would silently rename the unknowns.

c)

Augmented matrix:

(44452−1−2417−2−34111−2−3−30)\displaystyle \left(\begin{array}{ccc|c} 4 & 4 & 4 & 52 \\ -1 & -2 & 4 & 17 \\ -2 & -3 & 4 & 11 \\ 1 & -2 & -3 & -30 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+4R1R_{2} \to R_{2} + 4R_{1}, R3→R3−2R1R_{3} \to R_{3} - 2R_{1}, R4→R4+R1R_{4} \to R_{4} + R_{1}, R2→14R2R_{2} \to \frac{1}{4}R_{2}.

(−1−24170−153001−4−230−41−13)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 4 & 17 \\ 0 & -1 & 5 & 30 \\ 0 & 1 & -4 & -23 \\ 0 & -4 & 1 & -13 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R4→R4−4R2R_{4} \to R_{4} - 4R_{2}, R4→119R4R_{4} \to \frac{1}{19}R_{4}.

(−1−24170−1530001700−1−7)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 4 & 17 \\ 0 & -1 & 5 & 30 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & -1 & -7 \end{array}\right)

Column 33: R4→R4+R3R_{4} \to R_{4} + R_{3}.

(−1−24170−153000170000)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 4 & 17 \\ 0 & -1 & 5 & 30 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 \end{array}\right)

Row 44 reads 0=00 = 0: no contradiction, the system is consistent.

Row 33: x3=7x_{3} = 7.

Row 22: −x2+5x3=30-x_{2} + 5x_{3} = 30, with x3=7x_{3} = 7, gives x2=5x_{2} = 5.

Row 11: −x1−2x2+4x3=17-x_{1} - 2x_{2} + 4x_{3} = 17, with x2=5x_{2} = 5, x3=7x_{3} = 7, gives x1=1x_{1} = 1.

Check in the original system: 4+20+28=524 + 20 + 28 = 52, −1−10+28=17-1 - 10 + 28 = 17, −2−15+28=11-2 - 15 + 28 = 11, 1−10−21=−301 - 10 - 21 = -30.

Pitfall: four equations for three unknowns is not a reason to expect no solution. Only a row (0,…,0∣c)(0, \dots, 0 \mid c) with c≠0c \neq 0 proves that; here the extra row reduces to 0=00 = 0.

d)

Augmented matrix:

(41222−2−43−4−1224321428)\displaystyle \left(\begin{array}{cc|c} 4 & 1 & 22 \\ 2 & -2 & -4 \\ 3 & -4 & -12 \\ 2 & 4 & 32 \\ 1 & 4 & 28 \end{array}\right)

Column 11: R1↔R5R_{1} \leftrightarrow R_{5}, R2→R2−2R1R_{2} \to R_{2} - 2R_{1}, R3→R3−3R1R_{3} \to R_{3} - 3R_{1}, R4→R4−2R1R_{4} \to R_{4} - 2R_{1}, R5→R5−4R1R_{5} \to R_{5} - 4R_{1}, R2→110R2R_{2} \to \frac{1}{10}R_{2}, R3→116R3R_{3} \to \frac{1}{16}R_{3}, R4→14R4R_{4} \to \frac{1}{4}R_{4}, R5→115R5R_{5} \to \frac{1}{15}R_{5}.

(14280−1−60−1−60−1−60−1−6)\displaystyle \left(\begin{array}{cc|c} 1 & 4 & 28 \\ 0 & -1 & -6 \\ 0 & -1 & -6 \\ 0 & -1 & -6 \\ 0 & -1 & -6 \end{array}\right)

Column 22: R3→R3−R2R_{3} \to R_{3} - R_{2}, R4→R4−R2R_{4} \to R_{4} - R_{2}, R5→R5−R2R_{5} \to R_{5} - R_{2}.

(14280−1−6000000000)\displaystyle \left(\begin{array}{cc|c} 1 & 4 & 28 \\ 0 & -1 & -6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)

Rows 33, 44, 55 read 0=00 = 0: no contradiction, the system is consistent.

Row 22: −x2=−6-x_{2} = -6, so x2=6x_{2} = 6.

Row 11: x1+4x2=28x_{1} + 4x_{2} = 28, with x2=6x_{2} = 6, gives x1=4x_{1} = 4.

Check in the original system: 16+6=2216 + 6 = 22, 8−12=−48 - 12 = -4, 12−24=−1212 - 24 = -12, 8+24=328 + 24 = 32, 4+24=284 + 24 = 28.

Pitfall: solving with the first two equations and stopping. The other three must be carried to the end: here they all reduce to 0=00 = 0, so the point satisfies all five.

e)

Augmented matrix:

(3−4−3−20−112923172−3−1−7)\displaystyle \left(\begin{array}{ccc|c} 3 & -4 & -3 & -20 \\ -1 & 1 & 2 & 9 \\ 2 & 3 & 1 & 7 \\ 2 & -3 & -1 & -7 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+3R1R_{2} \to R_{2} + 3R_{1}, R3→R3+2R1R_{3} \to R_{3} + 2R_{1}, R4→R4+2R1R_{4} \to R_{4} + 2R_{1}, R3→15R3R_{3} \to \frac{1}{5}R_{3}.

(−11290−13701150−1311)\displaystyle \left(\begin{array}{ccc|c} -1 & 1 & 2 & 9 \\ 0 & -1 & 3 & 7 \\ 0 & 1 & 1 & 5 \\ 0 & -1 & 3 & 11 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R4→R4−R2R_{4} \to R_{4} - R_{2}, R3→14R3R_{3} \to \frac{1}{4}R_{3}, R4→14R4R_{4} \to \frac{1}{4}R_{4}.

(−11290−13700130001)\displaystyle \left(\begin{array}{ccc|c} -1 & 1 & 2 & 9 \\ 0 & -1 & 3 & 7 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right)

Row 44 reads 0=10 = 1: impossible, so the system has no solution.

Check without the matrix: with EiE_i the ii-th original equation, −E1−E2+E4-E_{1} - E_{2} + E_{4} cancels every unknown on the left and gives 44 on the right, so the equations contradict each other.

Pitfall: solving the first three equations, which do meet at exactly one point, (−1,2,3)(-1, 2, 3), and handing it in. A solution must satisfy EVERY equation, and the fourth gives 2(−1)−3(2)−3=−11≠−72(-1) - 3(2) - 3 = -11 \neq -7 at that point.

f)

Augmented matrix:

(−4−3010230−322−1022−12033−434)\displaystyle \left(\begin{array}{cccc|c} -4 & -3 & 0 & 1 & 0 \\ 2 & 3 & 0 & -3 & 22 \\ -1 & 0 & 2 & 2 & -12 \\ 0 & 3 & 3 & -4 & 34 \end{array}\right)

Column 11: R1↔R3R_{1} \leftrightarrow R_{3}, R2→R2+2R1R_{2} \to R_{2} + 2R_{1}, R3→R3−4R1R_{3} \to R_{3} - 4R_{1}.

(−1022−120341−20−3−8−748033−434)\displaystyle \left(\begin{array}{cccc|c} -1 & 0 & 2 & 2 & -12 \\ 0 & 3 & 4 & 1 & -2 \\ 0 & -3 & -8 & -7 & 48 \\ 0 & 3 & 3 & -4 & 34 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R4→R4−R2R_{4} \to R_{4} - R_{2}, R3→12R3R_{3} \to \frac{1}{2}R_{3}.

(−1022−120341−200−2−32300−1−536)\displaystyle \left(\begin{array}{cccc|c} -1 & 0 & 2 & 2 & -12 \\ 0 & 3 & 4 & 1 & -2 \\ 0 & 0 & -2 & -3 & 23 \\ 0 & 0 & -1 & -5 & 36 \end{array}\right)

Column 33: R3↔R4R_{3} \leftrightarrow R_{4}, R4→R4−2R3R_{4} \to R_{4} - 2R_{3}, R4→17R4R_{4} \to \frac{1}{7}R_{4}.

(−1022−120341−200−1−5360001−7)\displaystyle \left(\begin{array}{cccc|c} -1 & 0 & 2 & 2 & -12 \\ 0 & 3 & 4 & 1 & -2 \\ 0 & 0 & -1 & -5 & 36 \\ 0 & 0 & 0 & 1 & -7 \end{array}\right)

Row 44: x4=−7x_{4} = -7.

Row 33: −x3−5x4=36-x_{3} - 5x_{4} = 36, with x4=−7x_{4} = -7, gives x3=−1x_{3} = -1.

Row 22: 3x2+4x3+x4=−23x_{2} + 4x_{3} + x_{4} = -2, with x3=−1x_{3} = -1, x4=−7x_{4} = -7, gives x2=3x_{2} = 3.

Row 11: −x1+2x3+2x4=−12-x_{1} + 2x_{3} + 2x_{4} = -12, with x3=−1x_{3} = -1, x4=−7x_{4} = -7, gives x1=−4x_{1} = -4.

Check in the original system: 16−9−7=016 - 9 - 7 = 0, −8+9+21=22-8 + 9 + 21 = 22, 4−2−14=−124 - 2 - 14 = -12, 9−3+28=349 - 3 + 28 = 34.

Pitfall: a 00 in the system is a 00 in the matrix. Each equation misses one unknown, and copying its coefficients one column to the left would change the system.

g)

Augmented matrix:

(−113−127−323−245−13−3−39)\displaystyle \left(\begin{array}{cccc|c} -1 & 1 & 3 & -1 & 27 \\ -3 & 2 & 3 & -2 & 45 \\ -1 & 3 & -3 & -3 & 9 \end{array}\right)

Column 11: R2→R2−3R1R_{2} \to R_{2} - 3R_{1}, R3→R3−R1R_{3} \to R_{3} - R_{1}, R3→12R3R_{3} \to \frac{1}{2}R_{3}.

(−113−1270−1−61−3601−3−1−9)\displaystyle \left(\begin{array}{cccc|c} -1 & 1 & 3 & -1 & 27 \\ 0 & -1 & -6 & 1 & -36 \\ 0 & 1 & -3 & -1 & -9 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R3→19R3R_{3} \to \frac{1}{9}R_{3}.

(−113−1270−1−61−3600−10−5)\displaystyle \left(\begin{array}{cccc|c} -1 & 1 & 3 & -1 & 27 \\ 0 & -1 & -6 & 1 & -36 \\ 0 & 0 & -1 & 0 & -5 \end{array}\right)

Pivots in columns 11, 22, 33; x4x_{4} free, so x4=tx_{4} = t.

Row 33: −x3=−5-x_{3} = -5, so x3=5x_{3} = 5.

Row 22: −x2−6x3+x4=−36-x_{2} - 6x_{3} + x_{4} = -36, with x3=5x_{3} = 5, gives x2=6+tx_{2} = 6 + t.

Row 11: −x1+x2+3x3−x4=27-x_{1} + x_{2} + 3x_{3} - x_{4} = 27, with x2=6+tx_{2} = 6 + t, x3=5x_{3} = 5, gives x1=−6x_{1} = -6.

General solution, with tt any real number: (x1,x2,x3,x4)=(−6,6+t,5,t)(x_{1}, x_{2}, x_{3}, x_{4}) = (-6, 6 + t, 5, t).

Check with every parameter at 00, the point (−6,6,5,0)(-6, 6, 5, 0): 6+6+15+0=276 + 6 + 15 + 0 = 27, 18+12+15+0=4518 + 12 + 15 + 0 = 45, 6+18−15+0=96 + 18 - 15 + 0 = 9.

Pitfall: believing every variable depends on the parameter. Here x1x_1 and x3x_3 are fixed: every solution has x1=−6x_1 = -6 and x3=5x_3 = 5, and only x2x_2 moves with x4x_4.

h)

Augmented matrix:

(−4−2−440121−1−2−106301−3−1−2)\displaystyle \left(\begin{array}{ccccc|c} -4 & -2 & -4 & 4 & 0 & 12 \\ 1 & -1 & -2 & -1 & 0 & 6 \\ 3 & 0 & 1 & -3 & -1 & -2 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+4R1R_{2} \to R_{2} + 4R_{1}, R3→R3−3R1R_{3} \to R_{3} - 3R_{1}, R2→16R2R_{2} \to \frac{1}{6}R_{2}.

(1−1−2−1060−1−20060370−1−20)\displaystyle \left(\begin{array}{ccccc|c} 1 & -1 & -2 & -1 & 0 & 6 \\ 0 & -1 & -2 & 0 & 0 & 6 \\ 0 & 3 & 7 & 0 & -1 & -20 \end{array}\right)

Column 22: R3→R3+3R2R_{3} \to R_{3} + 3R_{2}.

(1−1−2−1060−1−20060010−1−2)\displaystyle \left(\begin{array}{ccccc|c} 1 & -1 & -2 & -1 & 0 & 6 \\ 0 & -1 & -2 & 0 & 0 & 6 \\ 0 & 0 & 1 & 0 & -1 & -2 \end{array}\right)

Pivots in columns 11, 22, 33; x4x_{4}, x5x_{5} free, so x4=sx_{4} = s, x5=tx_{5} = t.

Row 33: x3−x5=−2x_{3} - x_{5} = -2, so x3=−2+tx_{3} = -2 + t.

Row 22: −x2−2x3=6-x_{2} - 2x_{3} = 6, with x3=−2+tx_{3} = -2 + t, gives x2=−2−2tx_{2} = -2 - 2t.

Row 11: x1−x2−2x3−x4=6x_{1} - x_{2} - 2x_{3} - x_{4} = 6, with x2=−2−2tx_{2} = -2 - 2t, x3=−2+tx_{3} = -2 + t, gives x1=sx_{1} = s.

General solution, with ss, tt any real numbers: (x1,x2,x3,x4,x5)=(s,−2−2t,−2+t,s,t)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (s, -2 - 2t, -2 + t, s, t).

Check with every parameter at 00, the point (0,−2,−2,0,0)(0, -2, -2, 0, 0): 0+4+8+0=120 + 4 + 8 + 0 = 12, 0+2+4+0=60 + 2 + 4 + 0 = 6, 0−2+0+0=−20 - 2 + 0 + 0 = -2.

Pitfall: the column of x5x_5 is empty in the first two equations, which does not make x5x_5 zero, nor a pivot. It is free because its column holds no pivot.

i)

Augmented matrix:

(−2−14−312811−11−1−1−2)\displaystyle {\small \left(\begin{array}{cccccc|c} -2 & -1 & 4 & -3 & 1 & 2 & 8 \\ 1 & 1 & -1 & 1 & -1 & -1 & -2 \end{array}\right)}

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+2R1R_{2} \to R_{2} + 2R_{1}.

(11−11−1−1−2012−1−104)\displaystyle {\small \left(\begin{array}{cccccc|c} 1 & 1 & -1 & 1 & -1 & -1 & -2 \\ 0 & 1 & 2 & -1 & -1 & 0 & 4 \end{array}\right)}

Pivots in columns 11, 22; x3x_{3}, x4x_{4}, x5x_{5}, x6x_{6} free, so x3=rx_{3} = r, x4=sx_{4} = s, x5=tx_{5} = t, x6=ux_{6} = u.

Row 22: x2+2x3−x4−x5=4x_{2} + 2x_{3} - x_{4} - x_{5} = 4, so x2=4−2r+s+tx_{2} = 4 - 2r + s + t.

Row 11: x1+x2−x3+x4−x5−x6=−2x_{1} + x_{2} - x_{3} + x_{4} - x_{5} - x_{6} = -2, with x2=4−2r+s+tx_{2} = 4 - 2r + s + t, gives x1=−6+3r−2s+ux_{1} = -6 + 3r - 2s + u.

General solution, with rr, ss, tt, uu any real numbers: (x1,x2,x3,x4,x5,x6)=(−6+3r−2s+u,4−2r+s+t,r,s,t,u)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}) = (-6 + 3r - 2s + u, 4 - 2r + s + t, r, s, t, u).

Check with every parameter at 00, the point (−6,4,0,0,0,0)(-6, 4, 0, 0, 0, 0): 12−4+0+0+0+0=812 - 4 + 0 + 0 + 0 + 0 = 8, −6+4+0+0+0+0=−2-6 + 4 + 0 + 0 + 0 + 0 = -2.

Pitfall: counting the free variables from the shape of the system. Two equations in six unknowns leave AT LEAST four free variables; the exact number is read off the echelon form, here two pivots, so 6−2=46 - 2 = 4.

j)

Augmented matrix:

(1−2−3−40−2−61−1−2−210−320−32−31−10)\displaystyle {\small \left(\begin{array}{cccccc|c} 1 & -2 & -3 & -4 & 0 & -2 & -6 \\ 1 & -1 & -2 & -2 & 1 & 0 & -3 \\ 2 & 0 & -3 & 2 & -3 & 1 & -10 \end{array}\right)}

Column 11: R2→R2−R1R_{2} \to R_{2} - R_{1}, R3→R3−2R1R_{3} \to R_{3} - 2R_{1}.

(1−2−3−40−2−6011212304310−352)\displaystyle {\small \left(\begin{array}{cccccc|c} 1 & -2 & -3 & -4 & 0 & -2 & -6 \\ 0 & 1 & 1 & 2 & 1 & 2 & 3 \\ 0 & 4 & 3 & 10 & -3 & 5 & 2 \end{array}\right)}

Column 22: R3→R3−4R2R_{3} \to R_{3} - 4R_{2}.

(1−2−3−40−2−6011212300−12−7−3−10)\displaystyle {\small \left(\begin{array}{cccccc|c} 1 & -2 & -3 & -4 & 0 & -2 & -6 \\ 0 & 1 & 1 & 2 & 1 & 2 & 3 \\ 0 & 0 & -1 & 2 & -7 & -3 & -10 \end{array}\right)}

Pivots in columns 11, 22, 33; x4x_{4}, x5x_{5}, x6x_{6} free, so x4=rx_{4} = r, x5=sx_{5} = s, x6=tx_{6} = t.

Row 33: −x3+2x4−7x5−3x6=−10-x_{3} + 2x_{4} - 7x_{5} - 3x_{6} = -10, so x3=10+2r−7s−3tx_{3} = 10 + 2r - 7s - 3t.

Row 22: x2+x3+2x4+x5+2x6=3x_{2} + x_{3} + 2x_{4} + x_{5} + 2x_{6} = 3, with x3=10+2r−7s−3tx_{3} = 10 + 2r - 7s - 3t, gives x2=−7−4r+6s+tx_{2} = -7 - 4r + 6s + t.

Row 11: x1−2x2−3x3−4x4−2x6=−6x_{1} - 2x_{2} - 3x_{3} - 4x_{4} - 2x_{6} = -6, with x2=−7−4r+6s+tx_{2} = -7 - 4r + 6s + t, x3=10+2r−7s−3tx_{3} = 10 + 2r - 7s - 3t, gives x1=10+2r−9s−5tx_{1} = 10 + 2r - 9s - 5t.

General solution, with rr, ss, tt any real numbers: (x1,x2,x3,x4,x5,x6)=(10+2r−9s−5t,−7−4r+6s+t,10+2r−7s−3t,r,s,t)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}) = (10 + 2r - 9s - 5t, -7 - 4r + 6s + t, 10 + 2r - 7s - 3t, r, s, t).

Check with every parameter at 00, the point (10,−7,10,0,0,0)(10, -7, 10, 0, 0, 0): 10+14−30+0+0=−610 + 14 - 30 + 0 + 0 = -6, 10+7−20+0+0=−310 + 7 - 20 + 0 + 0 = -3, 20−30+0+0+0=−1020 - 30 + 0 + 0 + 0 = -10.

Pitfall: taking a pivot variable as a parameter. The pivots sit in columns 11, 22 and 33, so the parameters are x4x_4, x5x_5 and x6x_6, and nothing else.

k)

Augmented matrix:

(3−2−1−10−1−2−128−242−1−12)\displaystyle \left(\begin{array}{cccc|c} 3 & -2 & -1 & -1 & 0 \\ -1 & -2 & -1 & 2 & 8 \\ -2 & 4 & 2 & -1 & -12 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+3R1R_{2} \to R_{2} + 3R_{1}, R3→R3−2R1R_{3} \to R_{3} - 2R_{1}.

(−1−2−1280−8−4524084−5−28)\displaystyle \left(\begin{array}{cccc|c} -1 & -2 & -1 & 2 & 8 \\ 0 & -8 & -4 & 5 & 24 \\ 0 & 8 & 4 & -5 & -28 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R3→14R3R_{3} \to \frac{1}{4}R_{3}.

(−1−2−1280−8−45240000−1)\displaystyle \left(\begin{array}{cccc|c} -1 & -2 & -1 & 2 & 8 \\ 0 & -8 & -4 & 5 & 24 \\ 0 & 0 & 0 & 0 & -1 \end{array}\right)

Row 33 reads 0=−10 = -1: impossible, so the system has no solution.

Check without the matrix: with EiE_i the ii-th original equation, E1+E2+E3E_{1} + E_{2} + E_{3} cancels every unknown on the left and gives −4-4 on the right, so the equations contradict each other.

Pitfall: more unknowns than equations does not guarantee a solution. It only rules out a UNIQUE solution; the system can still have none, as here.

l)

Augmented matrix:

(440−20−1−2012−20420−2−301734−2−34)\displaystyle \left(\begin{array}{ccc|c} 4 & 4 & 0 & -20 \\ -1 & -2 & 0 & 12 \\ -2 & 0 & 4 & 20 \\ -2 & -3 & 0 & 17 \\ 3 & 4 & -2 & -34 \end{array}\right)

Column 11: R1↔R2R_{1} \leftrightarrow R_{2}, R2→R2+4R1R_{2} \to R_{2} + 4R_{1}, R3→R3−2R1R_{3} \to R_{3} - 2R_{1}, R4→R4−2R1R_{4} \to R_{4} - 2R_{1}, R5→R5+3R1R_{5} \to R_{5} + 3R_{1}, R2→14R2R_{2} \to \frac{1}{4}R_{2}, R3→14R3R_{3} \to \frac{1}{4}R_{3}, R5→12R5R_{5} \to \frac{1}{2}R_{5}.

(−1−20120−107011−1010−70−1−11)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 0 & 12 \\ 0 & -1 & 0 & 7 \\ 0 & 1 & 1 & -1 \\ 0 & 1 & 0 & -7 \\ 0 & -1 & -1 & 1 \end{array}\right)

Column 22: R3→R3+R2R_{3} \to R_{3} + R_{2}, R4→R4+R2R_{4} \to R_{4} + R_{2}, R5→R5−R2R_{5} \to R_{5} - R_{2}.

(−1−20120−1070016000000−1−6)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 0 & 12 \\ 0 & -1 & 0 & 7 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -6 \end{array}\right)

Column 33: R5→R5+R3R_{5} \to R_{5} + R_{3}.

(−1−20120−107001600000000)\displaystyle \left(\begin{array}{ccc|c} -1 & -2 & 0 & 12 \\ 0 & -1 & 0 & 7 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)

Rows 44, 55 read 0=00 = 0: no contradiction, the system is consistent.

Row 33: x3=6x_{3} = 6.

Row 22: −x2=7-x_{2} = 7, so x2=−7x_{2} = -7.

Row 11: −x1−2x2=12-x_{1} - 2x_{2} = 12, with x2=−7x_{2} = -7, gives x1=2x_{1} = 2.

Check in the original system: 8−28=−208 - 28 = -20, −2+14=12-2 + 14 = 12, −4+24=20-4 + 24 = 20, −4+21=17-4 + 21 = 17, 6−28−12=−346 - 28 - 12 = -34.

Pitfall: four of the five equations miss an unknown, and the missing term must be written as a 00 in its own column before any operation.

Control: a solution is substituted in EVERY original equation, never only in the echelon form, where an error made on the way would go unseen. A general solution is checked for at least one value of the parameters; a contradiction is confirmed by the combination of the original equations that produces it.

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Part A: the basics (/50)

Exercise 1: From equations to the augmented matrix, and which moves are legal

16 answers to check

Before any elimination, the system has to be copied into a matrix, and this is where the first marks are lost: a variable out of order, a missing variable with no 00 written in its place, a constant left on the wrong side. Then comes the only rule of the chapter: a row operation is legal when it can be undone, because only then does the new system have exactly the same solutions as the old one.

Consider the system {2y−x+3z=7z=5−2x3x−y=2z+1\begin{cases} 2y - x + 3z = 7 \\ z = 5 - 2x \\ 3x - y = 2z + 1 \end{cases} in the unknowns x,y,zx, y, z, taken in that order.

The figure belongs to part e): it shows the lines x+y=4x + y = 4 and x−y=−2x - y = -2, and the line that replaces the second one after the operation R2→R2−R1R_2 \to R_2 - R_1.

−1123456−1123456x + y = 4x - y = -2new R2: y = 3(1, 3)
  • a) Write the augmented matrix of the system, columns in the order x,y,zx, y, z.
  • b) Four operations are proposed on that matrix:(i)R1↔R3R_1 \leftrightarrow R_3(ii)R2→3R2−2R3R_2 \to 3R_2 - 2R_3(iii)R2→R3−R1R_2 \to R_3 - R_1(iv)R3→R3+3R1R_3 \to R_3 + 3R_1.Say which are legal, and name the elementary operations behind each legal one.
  • c) Solve the system by Gaussian elimination, naming every row operation, and check the answer in the ORIGINAL equations.
  • d) A student applies (iii) and carries on. Show that (3,−4,6)(3, -4, 6) satisfies the system he now works with but not the original one, and say what went wrong.
  • e) On the figure, explain why the new line produced by R2→R2−R1R_2 \to R_2 - R_1 passes through the crossing point of the other two, and why no new crossing point can appear.

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a)
b)
c)
d)
e)
Show the solution

Answers

  • a) (−123720153−1−21)\left(\begin{array}{ccc|c} -1 & 2 & 3 & 7 \\ 2 & 0 & 1 & 5 \\ 3 & -1 & -2 & 1 \end{array}\right)
  • b) (i), (ii) and (iv) are legal; (iii) is illegal, it deletes R2R_2.
  • c) (x,y,z)=(2,3,1)(x, y, z) = (2, 3, 1)
  • d) (3,−4,6)(3, -4, 6) solves the damaged system but gives 2x+z=12≠52x + z = 12 \neq 5: one equation was lost.
  • e) Any combination of the two equations holds at (1,3)(1, 3), and the move is undone by R2→R2+R1R_2 \to R_2 + R_1.
a)

Every equation is rewritten as (coefficient of xx) ++ (coefficient of yy) ++ (coefficient of zz) == constant BEFORE a single entry is copied. The first equation reads −x+2y+3z=7-x + 2y + 3z = 7: its terms were out of order, and copying them as written would have put the 22 in the xx column. The second, z=5−2xz = 5 - 2x, becomes 2x+0y+z=52x + 0y + z = 5: the variable yy is absent, and the absence is written as a 00, never skipped, otherwise every entry to its right slides one column to the left. The third, 3x−y=2z+13x - y = 2z + 1, becomes 3x−y−2z=13x - y - 2z = 1: the 2z2z changes sign as it crosses the equals sign. The augmented matrix is (−123720153−1−21)\left(\begin{array}{ccc|c} -1 & 2 & 3 & 7 \\ 2 & 0 & 1 & 5 \\ 3 & -1 & -2 & 1 \end{array}\right). The bar is not decoration: everything to its left is a coefficient, the last column holds the right-hand sides, and a row operation acts on the whole row, bar included. A column is a VARIABLE, never a position in the sentence.

b)

(i) is a swap, the first elementary operation, undone by swapping again: legal. (ii) is legal: it multiplies R2R_2 by 33, then adds −2R3-2R_3 to it, two elementary operations in a row. The row being REPLACED keeps a non-zero coefficient, 33, so the step is undone by R2→13(R2+2R3)R_2 \to \frac{1}{3}(R_2 + 2R_3). (iv) is the third elementary operation, adding a multiple of another row, undone by R3→R3−3R1R_3 \to R_3 - 3R_1: legal. (iii) is ILLEGAL. The new second row R3−R1R_3 - R_1 does not contain R2R_2 at all, so the information carried by the equation 2x+z=52x + z = 5 is thrown away and nothing left in the matrix can bring it back. The test that decides every case, including the ones with a parameter later in this set: the row being replaced must appear in its own replacement with a NON-ZERO coefficient. The multiplier of the OTHER row, on the contrary, can be anything, even 00.

c)

Make the leading entry of R1R_1 equal to 11: R1→−R1R_1 \to -R_1 gives (1,−2,−3∣−7)(1, -2, -3 \mid -7). Clear the first column: R2→R2−2R1R_2 \to R_2 - 2R_1 gives (0,4,7∣19)(0, 4, 7 \mid 19) and R3→R3−3R1R_3 \to R_3 - 3R_1 gives (0,5,7∣22)(0, 5, 7 \mid 22). Rather than dividing by 44 and meeting quarters, use R3→R3−R2R_3 \to R_3 - R_2, which gives (0,1,0∣3)(0, 1, 0 \mid 3), then R2↔R3R_2 \leftrightarrow R_3 to bring that convenient row up. Now R3→R3−4R2R_3 \to R_3 - 4R_2 gives (0,0,7∣7)(0, 0, 7 \mid 7) and R3→17R3R_3 \to \frac{1}{7}R_3 gives (0,0,1∣1)(0, 0, 1 \mid 1). The echelon form is (1−2−3−701030011)\left(\begin{array}{ccc|c} 1 & -2 & -3 & -7 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}\right). Back substitution: z=1z = 1, y=3y = 3, and x−2y−3z=−7x - 2y - 3z = -7 gives x=−7+6+3=2x = -7 + 6 + 3 = 2. Check in the ORIGINAL equations: 2(3)−2+3(1)=72(3) - 2 + 3(1) = 7, 1=5−2(2)1 = 5 - 2(2), 3(2)−3=2(1)+13(2) - 3 = 2(1) + 1. All three hold, so (x,y,z)=(2,3,1)(x, y, z) = (2, 3, 1). A check in the echelon form would only test the back substitution; the original system is the only one that also catches a wrong row operation, which is exactly what part d) is about.

d)

After (iii) the student works with R1R_1, R3−R1R_3 - R_1 and R3R_3, that is −x+2y+3z=7-x + 2y + 3z = 7, 4x−3y−5z=−64x - 3y - 5z = -6 and 3x−y−2z=13x - y - 2z = 1. For (3,−4,6)(3, -4, 6): −3−8+18=7-3 - 8 + 18 = 7, 12+12−30=−612 + 12 - 30 = -6 and 9+4−12=19 + 4 - 12 = 1, so it satisfies all three. But in the original second equation, 2x+z=6+6=12≠52x + z = 6 + 6 = 12 \neq 5: it is not a solution of the real system. What went wrong: the middle equation of the new system is the difference of the other two, so it says nothing new, and the student is really solving two equations in three unknowns. The solution set has grown from one point to a whole line through (2,3,1)(2, 3, 1) and (3,−4,6)(3, -4, 6). Notice the trap inside the trap: the true solution (2,3,1)(2, 3, 1) survives the illegal move, so a student who only checks the answer he expected sees nothing wrong. An illegal operation rarely loses the solution; it lets false ones in.

e)

Take the crossing point: it satisfies x+y=4x + y = 4 and x−y=−2x - y = -2, so it also satisfies their difference, (x−y)−(x+y)=−2−4(x - y) - (x + y) = -2 - 4, that is −2y=−6-2y = -6, or y=3y = 3. The crossing point (1,3)(1, 3) therefore lies on the new line, and the argument works whatever multiple of R1R_1 is subtracted: every operation of the third kind swings the line of R2R_2 around the solution point like a door on its hinge. No new crossing point can appear because the move can be undone: R2→R2+R1R_2 \to R_2 + R_1 brings the old line back, so a point lying on R1R_1 and on the new line also lies on the old one. The picture is the whole theory of the chapter in one drawing: the lines change at every step, the solution never does, and elimination is nothing more than turning the lines until they are horizontal and vertical, where the solution can be read.

Exercise 2: Two echelon forms, one reduced form

9 answers to check

Two students reduce the same augmented matrix (24−22131635−40)\left(\begin{array}{ccc|c} 2 & 4 & -2 & 2 \\ 1 & 3 & 1 & 6 \\ 3 & 5 & -4 & 0 \end{array}\right) and hand in different echelon forms. Both claim theirs is right. This exercise settles who is, and what exactly is guaranteed to be the same at the end of any correct reduction.

  • a) Student A starts with R1→12R1R_1 \to \frac{1}{2}R_1. Continue her reduction to a row echelon form, naming every operation.
  • b) Student B starts with R1↔R2R_1 \leftrightarrow R_2. Continue his reduction to a row echelon form, naming every operation.
  • c) The two echelon forms differ. Is one of them wrong? Say what the two forms must have in common, and what they need not.
  • d) Take both forms to the reduced row echelon form and compare.
  • e) Give the solution of the system, and explain why the reduced form can settle the question “is my work right?” when an echelon form cannot.

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c)
d)
e)
Show the solution

Answers

  • a) (12−1101250012)\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 2 \end{array}\right)
  • b) (13160−2−4−100012)\left(\begin{array}{ccc|c} 1 & 3 & 1 & 6 \\ 0 & -2 & -4 & -10 \\ 0 & 0 & 1 & 2 \end{array}\right)
  • c) Both are right: same pivot columns (11, 22, 33), different entries and different pivot values.
  • d) Both give (100101010012)\left(\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right).
  • e) (x,y,z)=(1,1,2)(x, y, z) = (1, 1, 2); the reduced form of a matrix is unique.
a)

R1→12R1R_1 \to \frac{1}{2}R_1 gives (1,2,−1∣1)(1, 2, -1 \mid 1). Then R2→R2−R1R_2 \to R_2 - R_1 gives (0,1,2∣5)(0, 1, 2 \mid 5) and R3→R3−3R1R_3 \to R_3 - 3R_1 gives (0,−1,−1∣−3)(0, -1, -1 \mid -3). Finally R3→R3+R2R_3 \to R_3 + R_2 gives (0,0,1∣2)(0, 0, 1 \mid 2). Her echelon form is (12−1101250012)\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 2 \end{array}\right). Dividing the first row by 22 was a good opening because every entry of that row is even: no fraction appears anywhere, and on a MATH 133 exam, where there is no calculator, choosing moves that keep integers is a genuine skill, not a matter of taste.

b)

R1↔R2R_1 \leftrightarrow R_2 puts (1,3,1∣6)(1, 3, 1 \mid 6) on top, with a leading 11 for free. Then R2→R2−2R1R_2 \to R_2 - 2R_1 gives (2,4,−2∣2)−(2,6,2∣12)=(0,−2,−4∣−10)(2, 4, -2 \mid 2) - (2, 6, 2 \mid 12) = (0, -2, -4 \mid -10) and R3→R3−3R1R_3 \to R_3 - 3R_1 gives (3,5,−4∣0)−(3,9,3∣18)=(0,−4,−7∣−18)(3, 5, -4 \mid 0) - (3, 9, 3 \mid 18) = (0, -4, -7 \mid -18). Finally R3→R3−2R2R_3 \to R_3 - 2R_2 gives (0,0,1∣2)(0, 0, 1 \mid 2). His echelon form is (13160−2−4−100012)\left(\begin{array}{ccc|c} 1 & 3 & 1 & 6 \\ 0 & -2 & -4 & -10 \\ 0 & 0 & 1 & 2 \end{array}\right). It is a perfectly valid echelon form: zero entries below each leading entry, each leading entry to the right of the one above. Nothing in the definition asks the leading entries to be 11.

c)

Neither is wrong. An echelon form is NOT unique: the entries to the right of the pivots, and even the values of the pivots themselves, 11 for A and −2-2 for B in the second row, depend on the route taken. What cannot depend on the route is the POSITION of the pivots: every echelon form of a given matrix has its leading entries in the same columns, here columns 11, 22 and 33, and therefore the same number of them. That is why the verdict, no solution, one, or infinitely many, can be read from any echelon form: it only uses where the pivots are. A student who compares his echelon form to the answer key entry by entry and concludes he is wrong has drawn the wrong conclusion; a student who compares pivot columns has not.

d)

From A: R2→R2−2R3R_2 \to R_2 - 2R_3 gives (0,1,0∣1)(0, 1, 0 \mid 1), R1→R1+R3R_1 \to R_1 + R_3 gives (1,2,0∣3)(1, 2, 0 \mid 3), then R1→R1−2R2R_1 \to R_1 - 2R_2 gives (1,0,0∣1)(1, 0, 0 \mid 1). From B: R2→−12R2R_2 \to -\frac{1}{2}R_2 gives (0,1,2∣5)(0, 1, 2 \mid 5), R2→R2−2R3R_2 \to R_2 - 2R_3 gives (0,1,0∣1)(0, 1, 0 \mid 1), R1→R1−R3R_1 \to R_1 - R_3 gives (1,3,0∣4)(1, 3, 0 \mid 4), then R1→R1−3R2R_1 \to R_1 - 3R_2 gives (1,0,0∣1)(1, 0, 0 \mid 1). Both routes end on (100101010012)\left(\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right), entry for entry. Working from the bottom up, clearing each column above its pivot, is the Gauss-Jordan half of the method, and it is shorter than it looks because the rows below are already clean.

e)

The reduced form reads the solution directly: (x,y,z)=(1,1,2)(x, y, z) = (1, 1, 2). Check in the original system: 2+4−4=22 + 4 - 4 = 2, 1+3+2=61 + 3 + 2 = 6, 3+5−8=03 + 5 - 8 = 0. The reason the reduced form settles disagreements is a theorem of the chapter: every matrix is row-equivalent to exactly ONE reduced row echelon form. Two correct reductions, however different their routes, must meet there; two reductions that end on different reduced forms cannot both be correct. So when your echelon form differs from a classmate's, do not argue about it: push both to the reduced form, or at least compare the pivot columns and the final solution. The echelon form is a stage on the way, the reduced form is the destination, and only the destination is the same for everyone.

Exercise 3: Pivot columns, free columns and the general solution

14 answers to check

When a column of the echelon form carries no pivot, its variable is free: it can take any value, and the pivot variables are then forced. The general solution lists all the solutions at once, with one parameter per free column, no more and no fewer.

Consider {x1+2x2−x4=32x1+4x2+x3=53x1+6x2+x3−x4=8\begin{cases} x_1 + 2x_2 - x_4 = 3 \\ 2x_1 + 4x_2 + x_3 = 5 \\ 3x_1 + 6x_2 + x_3 - x_4 = 8 \end{cases}.

  • a) Reduce the augmented matrix to reduced row echelon form, naming every operation.
  • b) Name the pivot variables and the free variables. Is x2x_2 free even though it sits between two pivot columns?
  • c) Write the general solution in parametric form, then as one vector plus multiples of fixed vectors.
  • d) Find the solution with x1=0x_1 = 0 and x2=0x_2 = 0.
  • e) Two classmates hand in: Ana, (5,−1,−1,0)+s(−2,1,0,0)+t(1,0,−2,1)(5, -1, -1, 0) + s(-2, 1, 0, 0) + t(1, 0, -2, 1); Ben, x1=3−2r+tx_1 = 3 - 2r + t, x2=rx_2 = r, x3=sx_3 = s, x4=tx_4 = t, with three parameters. Decide who is right, and prove it.

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  • a) (120−130012−100000)\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 3 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)
  • b) Pivots x1x_1, x3x_3; free x2x_2, x4x_4. Yes, x2x_2 is free.
  • c) (x1,x2,x3,x4)=(3,0,−1,0)+s(−2,1,0,0)+t(1,0,−2,1)(x_1, x_2, x_3, x_4) = (3, 0, -1, 0) + s(-2, 1, 0, 0) + t(1, 0, -2, 1), s,t∈Rs, t \in \mathbb{R}
  • d) (0,0,5,−3)(0, 0, 5, -3)
  • e) Ana is right (same set, shifted particular solution); Ben is wrong, (3,0,0,0)(3, 0, 0, 0) is on his list and fails.
a)

The augmented matrix is (120−1324105361−18)\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 3 \\ 2 & 4 & 1 & 0 & 5 \\ 3 & 6 & 1 & -1 & 8 \end{array}\right). R2→R2−2R1R_2 \to R_2 - 2R_1 gives (0,0,1,2∣−1)(0, 0, 1, 2 \mid -1) and R3→R3−3R1R_3 \to R_3 - 3R_1 gives (0,0,1,2∣−1)(0, 0, 1, 2 \mid -1). The second column was cleared at the same time as the first, so the next pivot jumps straight to column 33. Then R3→R3−R2R_3 \to R_3 - R_2 gives a row of zeros. The entry above the pivot of column 33 is already 00, so the reduced form is (120−130012−100000)\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 3 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right). The zero row appears because the third equation is the sum of the first two; it contains no 0=c0 = c with c≠0c \neq 0, so the system is consistent.

b)

The pivots sit in columns 11 and 33: x1x_1 and x3x_3 are the pivot variables. Columns 22 and 44 hold no pivot: x2x_2 and x4x_4 are free. Yes, x2x_2 is free even though it is not the last variable. Being free is decided by the ABSENCE of a pivot in the column, never by the position of the column. The jump of the second pivot from column 11 to column 33 is exactly what created a free variable in the middle. A frequent wrong answer declares only x4x_4 free, because free variables are expected at the end; it then produces a general solution with one parameter too few, which misses most of the solutions.

c)

Give each free variable a parameter, x2=sx_2 = s and x4=tx_4 = t, then solve each row of the REDUCED form for its pivot variable: x1=3−2s+tx_1 = 3 - 2s + t and x3=−1−2tx_3 = -1 - 2t. The general solution is (x1,x2,x3,x4)=(3−2s+t,s,−1−2t,t)(x_1, x_2, x_3, x_4) = (3 - 2s + t, s, -1 - 2t, t), with ss and tt any real numbers, or in vector form (3,0,−1,0)+s(−2,1,0,0)+t(1,0,−2,1)(3, 0, -1, 0) + s(-2, 1, 0, 0) + t(1, 0, -2, 1). Check the fixed vector in the original system: 3+0−0=33 + 0 - 0 = 3, 6+0−1=56 + 0 - 1 = 5, 9+0−1−0=89 + 0 - 1 - 0 = 8. The words “with ss and tt any real numbers” are part of the answer: without them the letters are unknowns and the line says nothing.

d)

x2=0x_2 = 0 means s=0s = 0. Then x1=3+t=0x_1 = 3 + t = 0 forces t=−3t = -3, so x4=−3x_4 = -3 and x3=−1−2(−3)=5x_3 = -1 - 2(-3) = 5. The solution is (0,0,5,−3)(0, 0, 5, -3). Check in the original system: 0+0−(−3)=30 + 0 - (-3) = 3, 0+0+5=50 + 0 + 5 = 5, 0+0+5−(−3)=80 + 0 + 5 - (-3) = 8. This is the everyday use of a general solution: once it is written, any extra condition on the variables becomes a small system in the parameters, here two equations in ss and tt.

e)

Ana is right. Her fixed vector (5,−1,−1,0)(5, -1, -1, 0) is a solution: 5−2−0=35 - 2 - 0 = 3, 10−4−1=510 - 4 - 1 = 5, 15−6−1−0=815 - 6 - 1 - 0 = 8; it is ours with s=−1s = -1. Her two direction vectors are ours, so as ss and tt run over all reals her list and ours describe the same set, only the starting point differs. Two correct general solutions can look different, and the way to compare them is substitution, never visual resemblance. Ben is wrong. He has three parameters but there are only two free columns: he treated x3x_3 as free although its column holds a pivot, so x3x_3 is not free, it is forced to equal −1−2x4-1 - 2x_4. His list contains points that are not solutions: r=s=t=0r = s = t = 0 gives (3,0,0,0)(3, 0, 0, 0), and the second equation gives 6+0+0=6≠56 + 0 + 0 = 6 \neq 5. Count before writing: the number of parameters equals the number of unknowns minus the number of pivots, here 4−2=24 - 2 = 2.

Exercise 4: A parameter k, first variant: the factor you must not cancel

11 answers to check

Systems with a parameter are the classic exam question of this chapter, and they are graded on one thing: whether every value of kk is accounted for. Consider {x+y+z=2x+2y−z=12x+3y+(k2−4)z=k+1\begin{cases} x + y + z = 2 \\ x + 2y - z = 1 \\ 2x + 3y + (k^2 - 4)z = k + 1 \end{cases}.

  • a) Reduce the augmented matrix to row echelon form, naming every operation.
  • b) Find the values of kk for which the system has no solution, exactly one solution, infinitely many solutions.
  • c) A student writes: “the last row reads (k−2)(k+2)z=k−2(k-2)(k+2)z = k - 2, so dividing by k−2k - 2, (k+2)z=1(k+2)z = 1 and z=1k+2z = \frac{1}{k+2}.” What has he lost, and what does his work claim for k=2k = 2?
  • d) Solve the system for k=2k = 2.
  • e) Give the unique solution in terms of kk when it exists, and check it for k=1k = 1 in the original system.

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  • a) (111201−2−100k2−4k−2)\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & k^2 - 4 & k - 2 \end{array}\right)
  • b) None for k=−2k = -2; infinitely many for k=2k = 2; exactly one for every other kk.
  • c) He lost the case k=2k = 2: his work claims one solution there, (94,−12,14)(\frac{9}{4}, -\frac{1}{2}, \frac{1}{4}), which is only one of infinitely many.
  • d) (x,y,z)=(3,−1,0)+t(−3,2,1)(x, y, z) = (3, -1, 0) + t(-3, 2, 1), t∈Rt \in \mathbb{R}
  • e) x=3k+3k+2x = \frac{3k+3}{k+2}, y=−kk+2y = -\frac{k}{k+2}, z=1k+2z = \frac{1}{k+2}; for k=1k = 1, (2,−13,13)(2, -\frac{1}{3}, \frac{1}{3}).
a)

Start from (111212−1123k2−4k+1)\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 1 & 2 & -1 & 1 \\ 2 & 3 & k^2 - 4 & k + 1 \end{array}\right). R2→R2−R1R_2 \to R_2 - R_1 gives (0,1,−2∣−1)(0, 1, -2 \mid -1). R3→R3−2R1R_3 \to R_3 - 2R_1 gives (0,1,k2−6∣k−3)(0, 1, k^2 - 6 \mid k - 3). Then R3→R3−R2R_3 \to R_3 - R_2 gives (0,0,k2−4∣k−2)(0, 0, k^2 - 4 \mid k - 2). The echelon form is (111201−2−100k2−4k−2)\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & k^2 - 4 & k - 2 \end{array}\right). Every multiplier used was a plain number, so every step is legal for every value of kk at once: no case has to be split yet, and that is the reason to push kk to the last row before doing anything with it.

b)

Factor the last pivot: k2−4=(k−2)(k+2)k^2 - 4 = (k - 2)(k + 2). It vanishes for k=2k = 2 and k=−2k = -2 only, so those are the only values that can be special, and each is studied separately in the echelon form. If k≠2k \neq 2 and k≠−2k \neq -2, the three columns hold a pivot, no variable is free and no row is contradictory: exactly one solution. If k=−2k = -2, the last row becomes (0,0,0∣−4)(0, 0, 0 \mid -4), that is 0=−40 = -4: no solution. If k=2k = 2, it becomes (0,0,0∣0)(0, 0, 0 \mid 0): the system is consistent and column 33 has lost its pivot, so zz is free: infinitely many solutions. The order of the reasoning is the method: first the values where the pivot vanishes, THEN the right-hand side at each of those values. The right-hand side k−2k - 2 is what separates k=2k = 2 from k=−2k = -2, although both kill the same pivot.

c)

Dividing by k−2k - 2 is multiplying by 1k−2\frac{1}{k - 2}, which does not exist when k=2k = 2. The step is illegal for exactly the value that matters, and after it the case k=2k = 2 has silently disappeared: his last row (k+2)z=1(k + 2)z = 1 still has a non-zero pivot at k=2k = 2, so his work claims ONE solution there, z=14z = \frac{1}{4}, y=−1+2z=−12y = -1 + 2z = -\frac{1}{2} and x=2−y−z=94x = 2 - y - z = \frac{9}{4}. The dangerous part is that this triple really is a solution for k=2k = 2, it is the case t=14t = \frac{1}{4} of part d), so substituting it back shows nothing wrong. What is wrong is the word “exactly”: he has found one point of a whole line. On a marking scheme the case k=2k = 2 is typically a third of the question, and it is lost entirely. Keep the factor, split the cases, and divide only inside the case where the divisor is known to be non-zero.

d)

For k=2k = 2 the echelon form is (111201−2−10000)\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right). Column 33 has no pivot, so z=tz = t with tt any real number. The second row gives y=−1+2ty = -1 + 2t, the first x=2−y−z=2+1−2t−t=3−3tx = 2 - y - z = 2 + 1 - 2t - t = 3 - 3t. The solutions are (x,y,z)=(3,−1,0)+t(−3,2,1)(x, y, z) = (3, -1, 0) + t(-3, 2, 1), a line in space. Check the third ORIGINAL equation at k=2k = 2, where it reads 2x+3y+0z=32x + 3y + 0z = 3: 2(3−3t)+3(−1+2t)=6−6t−3+6t=32(3 - 3t) + 3(-1 + 2t) = 6 - 6t - 3 + 6t = 3 for every tt, as it must.

e)

For k≠±2k \neq \pm 2 the division is now legal: z=k−2(k−2)(k+2)=1k+2z = \frac{k - 2}{(k - 2)(k + 2)} = \frac{1}{k + 2}. The second row gives y=−1+2z=−(k+2)+2k+2=−kk+2y = -1 + 2z = \frac{-(k + 2) + 2}{k + 2} = -\frac{k}{k + 2}, and the first x=2−y−z=2(k+2)+k−1k+2=3k+3k+2x = 2 - y - z = \frac{2(k + 2) + k - 1}{k + 2} = \frac{3k + 3}{k + 2}. For k=1k = 1: (x,y,z)=(2,−13,13)(x, y, z) = (2, -\frac{1}{3}, \frac{1}{3}). Check in the ORIGINAL system at k=1k = 1, where the third equation reads 2x+3y−3z=22x + 3y - 3z = 2: 2−13+13=22 - \frac{1}{3} + \frac{1}{3} = 2, 2−23−13=12 - \frac{2}{3} - \frac{1}{3} = 1, 4−1−1=24 - 1 - 1 = 2. All three hold. The formula also explains the picture of part b) from another side: it blows up at k=−2k = -2, where there is no solution, and at k=2k = 2 it gives a single point of the line of part d), which is exactly the student's mistake of part c).

-4-3-2-101234no solutioninfinitely manyevery other value of k: exactly one solutionk

Exercise 5: A parameter k, second variant: when k sits where the pivot should be

7 answers to check

Consider {kx+y+z=1x+ky+z=1x+y+kz=1\begin{cases} kx + y + z = 1 \\ x + ky + z = 1 \\ x + y + kz = 1 \end{cases}. The natural first move, dividing R1R_1 by kk to get a leading 11, is exactly the move this question is testing.

  • a) Explain why R1→1kR1R_1 \to \frac{1}{k}R_1 is not an allowed first step, and propose a first step that is legal for every kk.
  • b) Reduce to row echelon form, naming every operation, without ever dividing by an expression in kk.
  • c) Find the values of kk giving no solution, exactly one solution, infinitely many solutions.
  • d) For k=1k = 1, write the general solution. A student who divided by k−1k - 1 claims that for k=1k = 1 the only solution is x=y=z=13x = y = z = \frac{1}{3}. Is it a solution? Is it the only one?
  • e) Give the unique solution when it exists, and use it to find xx for k=0k = 0, the value the illegal first step would have excluded.

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  • a) It divides by 00 when k=0k = 0; swap instead, R1↔R3R_1 \leftrightarrow R_3.
  • b) (11k10k−11−k000(1−k)(k+2)1−k)\left(\begin{array}{ccc|c} 1 & 1 & k & 1 \\ 0 & k - 1 & 1 - k & 0 \\ 0 & 0 & (1 - k)(k + 2) & 1 - k \end{array}\right)
  • c) None for k=−2k = -2; infinitely many for k=1k = 1; exactly one otherwise.
  • d) (1−s−t,s,t)(1 - s - t, s, t); 13\frac{1}{3} each is a solution, not the only one: (1,0,0)(1, 0, 0) is another.
  • e) x=y=z=1k+2x = y = z = \frac{1}{k + 2}; for k=0k = 0, x=12x = \frac{1}{2}.
a)

R1→1kR1R_1 \to \frac{1}{k}R_1 multiplies a row by 1k\frac{1}{k}, which does not exist when k=0k = 0. The step would force a separate case k=0k = 0 before the reduction even starts, and a student who forgets that case has used an illegal operation. The fix costs nothing: R1↔R3R_1 \leftrightarrow R_3 brings (1,1,k∣1)(1, 1, k \mid 1) to the top, whose leading entry is 11 for every value of kk. The general rule: when a parameter sits where a pivot should be, look for a row whose leading entry is a NUMBER, and swap it up.

b)

After the swap, the rows are (1,1,k∣1)(1, 1, k \mid 1), (1,k,1∣1)(1, k, 1 \mid 1) and (k,1,1∣1)(k, 1, 1 \mid 1). R2→R2−R1R_2 \to R_2 - R_1 gives (0,k−1,1−k∣0)(0, k - 1, 1 - k \mid 0). R3→R3−kR1R_3 \to R_3 - kR_1 gives (0,1−k,1−k2∣1−k)(0, 1 - k, 1 - k^2 \mid 1 - k): this move is legal for every kk, even k=0k = 0, because the expression kk multiplies the OTHER row and the row being replaced keeps its coefficient 11. Then R3→R3+R2R_3 \to R_3 + R_2 gives (0,0,2−k−k2∣1−k)(0, 0, 2 - k - k^2 \mid 1 - k). Factor: 2−k−k2=(1−k)(k+2)2 - k - k^2 = (1 - k)(k + 2). The echelon form is (11k10k−11−k000(1−k)(k+2)1−k)\left(\begin{array}{ccc|c} 1 & 1 & k & 1 \\ 0 & k - 1 & 1 - k & 0 \\ 0 & 0 & (1 - k)(k + 2) & 1 - k \end{array}\right). Two leading entries depend on kk, k−1k - 1 and (1−k)(k+2)(1 - k)(k + 2), and none of them has been divided by.

c)

The pivots vanish only at k=1k = 1 and k=−2k = -2. If k≠1k \neq 1 and k≠−2k \neq -2, all three columns hold a pivot: exactly one solution. If k=−2k = -2, the second row is (0,−3,3∣0)(0, -3, 3 \mid 0), still a pivot, and the third is (0,0,0∣3)(0, 0, 0 \mid 3), that is 0=30 = 3: no solution. If k=1k = 1, the second and third rows are entirely zero: the system is the single equation x+y+z=1x + y + z = 1, consistent, with yy and zz free: infinitely many solutions. Geometrically, at k=1k = 1 the three equations are the same plane written three times, and the solution set is that whole plane.

d)

For k=1k = 1, put y=sy = s and z=tz = t: the general solution is (x,y,z)=(1−s−t,s,t)(x, y, z) = (1 - s - t, s, t), with s,ts, t any real numbers. The student's triple is a solution, 13+13+13=1\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1; it is the case s=t=13s = t = \frac{1}{3}. It is NOT the only one: (1,0,0)(1, 0, 0), (0,1,0)(0, 1, 0) and (2,−1,0)(2, -1, 0) are others. His division by k−1k - 1 turned the second row into y−z=0y - z = 0 and the third into (k+2)z=1(k + 2)z = 1, two equations that still have pivots at k=1k = 1, so his method could only ever produce one point of the plane. As in the previous exercise, the lost case does not give a false answer, it gives a TRUE answer with a false word, “only”, attached.

e)

For k≠1,−2k \neq 1, -2, dividing is now legal. The third row gives z=1−k(1−k)(k+2)=1k+2z = \frac{1 - k}{(1 - k)(k + 2)} = \frac{1}{k + 2}. The second, (k−1)(y−z)=0(k - 1)(y - z) = 0 with k−1≠0k - 1 \neq 0, gives y=zy = z. The first gives x=1−y−kz=1−1+kk+2=1k+2x = 1 - y - kz = 1 - \frac{1 + k}{k + 2} = \frac{1}{k + 2}. So x=y=z=1k+2x = y = z = \frac{1}{k + 2}, which the symmetry of the system predicted: exchanging the unknowns exchanges the equations. For k=0k = 0, x=y=z=12x = y = z = \frac{1}{2}; check: y+z=1y + z = 1, x+z=1x + z = 1, x+y=1x + y = 1. The value k=0k = 0, which the illegal first step would have singled out, is not special at all. That is the double cost of dividing by an expression in kk: it creates a fake case (k=0k = 0) and hides a real one (k=1k = 1).

Part B: problems and reasoning (/50)

Exercise 6: Three lines in the plane: a parameter that can only break the system

13 answers to check

Three equations in two unknowns are three lines in the plane, and a solution is a point lying on all three at once. Consider {x+2y=4x−y=1kx+y=5\begin{cases} x + 2y = 4 \\ x - y = 1 \\ kx + y = 5 \end{cases}. The figure draws the three lines for k=1k = 1.

−11234567−2−112345x + 2y = 4x - y = 1k = 1: x + y = 5
  • a) Reduce the augmented matrix to row echelon form, naming every operation. Where does kk end up?
  • b) Find the value of kk for which the system is consistent, and give the solution.
  • c) For k=1k = 1 (figure), find the three points where the lines cross two at a time, and explain why the system has no solution although every pair of lines meets.
  • d) For which values of kk is the third line parallel to one of the first two? How many solutions then?
  • e) Can this system have infinitely many solutions for some kk? Answer from the echelon form, then from the picture.

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  • a) (124011004−2k)\left(\begin{array}{cc|c} 1 & 2 & 4 \\ 0 & 1 & 1 \\ 0 & 0 & 4 - 2k \end{array}\right): kk ends up in the augmented column only.
  • b) k=2k = 2, solution (x,y)=(2,1)(x, y) = (2, 1).
  • c) (2,1)(2, 1), (6,−1)(6, -1), (3,2)(3, 2): three different points, a triangle, no common point.
  • d) k=12k = \frac{1}{2} (parallel to the first) and k=−1k = -1 (parallel to the second): no solution.
  • e) No: both variable columns always hold a pivot, and the first two lines are already different.
a)

Start from (1241−11k15)\left(\begin{array}{cc|c} 1 & 2 & 4 \\ 1 & -1 & 1 \\ k & 1 & 5 \end{array}\right). R2→R2−R1R_2 \to R_2 - R_1 gives (0,−3∣−3)(0, -3 \mid -3), and R2→−13R2R_2 \to -\frac{1}{3}R_2 gives (0,1∣1)(0, 1 \mid 1). R3→R3−kR1R_3 \to R_3 - kR_1 gives (0,1−2k∣5−4k)(0, 1 - 2k \mid 5 - 4k), legal for every kk since the expression multiplies the other row. Then R3→R3−(1−2k)R2R_3 \to R_3 - (1 - 2k)R_2 gives (0,0∣5−4k−(1−2k))=(0,0∣4−2k)(0, 0 \mid 5 - 4k - (1 - 2k)) = (0, 0 \mid 4 - 2k), legal for the same reason. The parameter has left the coefficient part entirely: it survives only in the augmented column. That already tells the whole story: kk cannot change the number of pivots among the variables, it can only decide whether the last row is a contradiction.

b)

The last row reads 0=4−2k0 = 4 - 2k. It is a contradiction unless 4−2k=04 - 2k = 0, that is k=2k = 2. For k=2k = 2 the rows give y=1y = 1, then x=4−2y=2x = 4 - 2y = 2: the solution is (2,1)(2, 1). Check the third equation: 2(2)+1=52(2) + 1 = 5. In the picture, k=2k = 2 is the unique slope for which the third line passes through the point where the first two already cross. A student who reasons “three equations, two unknowns, so no solution” gets every value but k=2k = 2 right and the only interesting one wrong: counting equations is never the test, the last row is.

c)

For k=1k = 1 the third line is x+y=5x + y = 5. First and second: subtracting gives 3y=33y = 3, so (2,1)(2, 1). First and third: subtracting gives y=−1y = -1, then x=6x = 6, so (6,−1)(6, -1). Second and third: adding gives 2x=62x = 6, so x=3x = 3, y=2y = 2, point (3,2)(3, 2). Every pair of lines meets, but at three DIFFERENT points, the vertices of the triangle on the figure. A solution of the system must lie on the three lines simultaneously, and no point does. The echelon form says the same thing in one line: at k=1k = 1 the last row reads 0=20 = 2.

d)

The third line kx+y=5kx + y = 5 has slope −k-k. The first, x+2y=4x + 2y = 4, has slope −12-\frac{1}{2}, and the second, x−y=1x - y = 1, has slope 11. So the third line is parallel to the first for k=12k = \frac{1}{2}, where it reads x+2y=10x + 2y = 10, and to the second for k=−1k = -1, where it reads x−y=−5x - y = -5. In both cases the lines are parallel and DISTINCT, so they never meet, and the system has no solution: the last row reads 0=30 = 3 and 0=60 = 6. The trap to avoid is the reflex “parallel means infinitely many”: that is only true for two lines that coincide, and here the right-hand sides differ.

e)

No. Whatever kk is, the echelon form keeps a pivot in both variable columns, 11 and 11, so no variable is ever free: the system has at most one solution, and exactly one when it is consistent. The picture agrees: infinitely many solutions would need the three lines to be one and the same line, and the first two, with slopes −12-\frac{1}{2} and 11, are different lines before kk enters the problem at all. So for this system the three textbook cases shrink to two: one solution for k=2k = 2, none otherwise. A parameter question does not promise all three cases, and inventing one to fill the pattern costs as much as missing one.

−1123456−2−112345x + 2y = 4x − y = 12x + y = 5(2, 1)

Exercise 7: Polynomial interpolation: a curve through given points

12 answers to check

Find the parabola y=a+bx+cx2y = a + bx + cx^2 that passes through (−1,8)(-1, 8), (1,2)(1, 2) and (2,5)(2, 5). The unknowns are not xx and yy: they are the coefficients aa, bb, cc, and each point gives one linear equation in them. The figure shows the three points and a fourth one, (3,10)(3, 10), used in part c).

−2−112342468101214(-1, 8)(1, 2)(2, 5)(3, 10)xy
  • a) Write the three equations in aa, bb, cc and the augmented matrix.
  • b) Reduce, naming every operation, give the parabola and check it at the three points.
  • c) The parabola is now also required to pass through (3,10)(3, 10). Add the fourth equation and show by row reduction that no parabola passes through the four points.
  • d) Show that infinitely many cubics y=a+bx+cx2+dx3y = a + bx + cx^2 + dx^3 pass through the three original points, give them all, and find the one that also passes through (3,10)(3, 10).
  • e) Two measurements are recorded as (1,2)(1, 2) and (1,3)(1, 3). What does elimination say about a parabola through both, and why?

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  • a) (1−11811121245)\left(\begin{array}{ccc|c} 1 & -1 & 1 & 8 \\ 1 & 1 & 1 & 2 \\ 1 & 2 & 4 & 5 \end{array}\right)
  • b) y=3−3x+2x2y = 3 - 3x + 2x^2
  • c) Last row (0,0,0∣−2)(0, 0, 0 \mid -2): inconsistent, the parabola gives 1212 at x=3x = 3, not 1010.
  • d) y=(3+2t)−(3+t)x+(2−2t)x2+tx3y = (3 + 2t) - (3 + t)x + (2 - 2t)x^2 + tx^3; through (3,10)(3, 10): t=−14t = -\frac{1}{4}.
  • e) Row (0,0,0∣1)(0, 0, 0 \mid 1): no parabola, no function at all, takes two values at x=1x = 1.
a)

Substituting each point into y=a+bx+cx2y = a + bx + cx^2: (−1,8)(-1, 8) gives a−b+c=8a - b + c = 8, (1,2)(1, 2) gives a+b+c=2a + b + c = 2, and (2,5)(2, 5) gives a+2b+4c=5a + 2b + 4c = 5. The augmented matrix is (1−11811121245)\left(\begin{array}{ccc|c} 1 & -1 & 1 & 8 \\ 1 & 1 & 1 & 2 \\ 1 & 2 & 4 & 5 \end{array}\right). The columns are the powers of xx, 11, xx, x2x^2, evaluated at each point; this is the first time in the course that the unknowns of a linear system are not coordinates, and it is exactly why these systems are linear even though the curve is not a line: aa, bb, cc appear to the first power only.

b)

R2→R2−R1R_2 \to R_2 - R_1 gives (0,2,0∣−6)(0, 2, 0 \mid -6) and R3→R3−R1R_3 \to R_3 - R_1 gives (0,3,3∣−3)(0, 3, 3 \mid -3). Then R2→12R2R_2 \to \frac{1}{2}R_2 gives (0,1,0∣−3)(0, 1, 0 \mid -3) and R3→13R3R_3 \to \frac{1}{3}R_3 gives (0,1,1∣−1)(0, 1, 1 \mid -1). Finally R3→R3−R2R_3 \to R_3 - R_2 gives (0,0,1∣2)(0, 0, 1 \mid 2). So c=2c = 2, b=−3b = -3, and a=8+b−c=3a = 8 + b - c = 3. The parabola is y=3−3x+2x2y = 3 - 3x + 2x^2. Check at the three points: 3+3+2=83 + 3 + 2 = 8, 3−3+2=23 - 3 + 2 = 2, 3−6+8=53 - 6 + 8 = 5. Three points with distinct xx-coordinates, three coefficients, three pivots: exactly one parabola.

c)

The point (3,10)(3, 10) adds the equation a+3b+9c=10a + 3b + 9c = 10, a fourth row (1,3,9∣10)(1, 3, 9 \mid 10). R4→R4−R1R_4 \to R_4 - R_1 gives (0,4,8∣2)(0, 4, 8 \mid 2); with the reduced rows of part b), R4→R4−4R2R_4 \to R_4 - 4R_2 gives (0,0,8∣14)(0, 0, 8 \mid 14), then R4→R4−8R3R_4 \to R_4 - 8R_3 gives (0,0,0∣−2)(0, 0, 0 \mid -2): 0=−20 = -2, the system is inconsistent. The meaning is concrete: the first three points already fix the parabola completely, and that parabola takes the value 3−9+18=123 - 9 + 18 = 12 at x=3x = 3, not 1010. The solution figure shows the miss. In a lab, this is the normal situation: more measurements than coefficients, no exact curve, which is why statistics courses replace “solve” by “fit”.

d)

With four coefficients the equations are a−b+c−d=8a - b + c - d = 8, a+b+c+d=2a + b + c + d = 2 and a+2b+4c+8d=5a + 2b + 4c + 8d = 5. R2→R2−R1R_2 \to R_2 - R_1 gives (0,2,0,2∣−6)(0, 2, 0, 2 \mid -6), then 12\frac{1}{2}: (0,1,0,1∣−3)(0, 1, 0, 1 \mid -3). R3→R3−R1R_3 \to R_3 - R_1 gives (0,3,3,9∣−3)(0, 3, 3, 9 \mid -3), then 13\frac{1}{3}: (0,1,1,3∣−1)(0, 1, 1, 3 \mid -1). R3→R3−R2R_3 \to R_3 - R_2 gives (0,0,1,2∣2)(0, 0, 1, 2 \mid 2). Pivots in the columns of aa, bb, cc; the column of dd is free. With d=td = t: c=2−2tc = 2 - 2t, b=−3−tb = -3 - t, a=8+b−c+d=3+2ta = 8 + b - c + d = 3 + 2t. All the cubics are y=(3+2t)−(3+t)x+(2−2t)x2+tx3y = (3 + 2t) - (3 + t)x + (2 - 2t)x^2 + tx^3, and t=0t = 0 gives back the parabola. At x=3x = 3 this is 12+8t12 + 8t, so passing through (3,10)(3, 10) requires 8t=−28t = -2, t=−14t = -\frac{1}{4}: four points, four coefficients, one cubic. The chapter's three cases in one exercise: three points and three coefficients, one curve; four points and three coefficients, none here; three points and four coefficients, infinitely many.

e)

Both points give the equation a+b+c=a + b + c = something: a+b+c=2a + b + c = 2 and a+b+c=3a + b + c = 3. R2→R2−R1R_2 \to R_2 - R_1 gives (0,0,0∣1)(0, 0, 0 \mid 1), that is 0=10 = 1: no parabola passes through both. No curve of the form y=f(x)y = f(x) can, since a function takes ONE value at x=1x = 1. Distinct xx-coordinates are the hidden hypothesis of every interpolation statement, and elimination detects its failure with the only tool it has for an impossibility, a pivot in the augmented column.

−2−112342468101214(-1, 8)(1, 2)(2, 5)(3, 10) missedp(3) = 12y = 3 - 3x + 2x²xy

Exercise 8: Five statements to correct

6 answers to check

Each statement below was written on a MATH 133 midterm, and each is false. Say what is wrong, give the correct statement, and give the smallest example that settles it.

  • a) “The reduced matrix has a row of zeros, so the system has infinitely many solutions.”
  • b) “My system has a free variable, so it has infinitely many solutions.”
  • c) “R1→R1−R2R_1 \to R_1 - R_2 and R2→R2−R1R_2 \to R_2 - R_1 can be done in the same step, since each one is elementary.”
  • d) “In (12030014)\left(\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 4 \end{array}\right) the free variable is x3x_3, because it comes last.”
  • e) “The last row is (0,0,(k−3)(k+1)∣2(k−3))(0, 0, (k - 3)(k + 1) \mid 2(k - 3)). Dividing by k−3k - 3, z=2k+1z = \frac{2}{k + 1}, so the system has exactly one solution for every k≠−1k \neq -1.”

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  • a) False: x=2x = 2, y=3y = 3, x+y=5x + y = 5 has a zero row and ONE solution.
  • b) False: x+y=1x + y = 1, 0=10 = 1 has yy free and no solution. Consistency first.
  • c) False: done on the old rows, the two new rows are opposites and an equation is lost.
  • d) False: x3x_3 has a pivot; the free variable is x2x_2, solution (3−2t,t,4)(3 - 2t, t, 4).
  • e) False: k=3k = 3 gives infinitely many, k=−1k = -1 none, the others exactly one.
a)

FALSE. A zero row only says that one equation was a consequence of the others; it says nothing about free variables. Smallest example: x=2x = 2, y=3y = 3, x+y=5x + y = 5 reduces to (102013000)\left(\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right), with a zero row and exactly one solution, (2,3)(2, 3). Correct statement: a CONSISTENT system has infinitely many solutions exactly when some variable column holds no pivot. Rows are counted to find contradictions, columns are counted to find freedom.

b)

FALSE, because it skips the first test. In (111001)\left(\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right) the column of yy has no pivot, so yy looks free, but the second row reads 0=10 = 1 and the system has no solution at all. Correct statement: IF the system is consistent and has a free variable, then it has infinitely many solutions. The reading of an echelon form has an order: first look for a row (0  ⋯  0∣c)(0 \; \cdots \; 0 \mid c) with c≠0c \neq 0, and only if there is none, count the free columns.

c)

FALSE. Each move is elementary on its own, but done together on the OLD rows they are not a sequence of elementary moves. Take x+y=3x + y = 3, x−y=1x - y = 1, whose only solution is (2,1)(2, 1). Simultaneously, the new R1R_1 is (0,2∣2)(0, 2 \mid 2) and the new R2R_2 is (0,−2∣−2)(0, -2 \mid -2), the opposite of the first: one equation has vanished, and (5,1)(5, 1) now satisfies the system although 5+1≠35 + 1 \neq 3. Correct statement: operations are performed one at a time, each on the CURRENT matrix. Done in sequence, R1→R1−R2R_1 \to R_1 - R_2 gives (0,2∣2)(0, 2 \mid 2), then R2→R2−R1R_2 \to R_2 - R_1 acts on that new row and gives (1,−3∣−1)(1, -3 \mid -1), and nothing is lost.

d)

FALSE. Free means “no pivot in the column”, not “last”. The pivots are in columns 11 and 33, so x1x_1 and x3x_3 are pivot variables and x2x_2 is the free one. The solution is x2=tx_2 = t, x1=3−2tx_1 = 3 - 2t, x3=4x_3 = 4: x3x_3 is not free at all, it is forced to be 44. Choosing x3x_3 as the parameter produces the nonsense x3=tx_3 = t and x3=4x_3 = 4 at the same time.

e)

FALSE: the division by k−3k - 3 is illegal at k=3k = 3, which is exactly where something happens. At k=3k = 3 the row is (0,0,0∣0)(0, 0, 0 \mid 0), so zz is free and, provided the rows above are consistent, there are infinitely many solutions. At k=−1k = -1 the row is (0,0,0∣−8)(0, 0, 0 \mid -8): no solution. Correct statement: exactly one solution for k≠3k \neq 3 and k≠−1k \neq -1, none for k=−1k = -1, infinitely many for k=3k = 3. The student's formula even gives a value at k=3k = 3, z=12z = \frac{1}{2}, which is one of the infinitely many solutions: true number, false word “exactly”.

Exercise 9: A circuit solved with Kirchhoff's laws

11 answers to check

Kirchhoff's junction law: at a junction, the current flowing in equals the current flowing out. Kirchhoff's loop law: around any closed loop, the sum of the battery voltages equals the sum of the drops RIRI across the resistors, each counted with the sign given by the direction chosen for its current.

In the circuit of the figure, a 99 V battery and a 1 Ω1\,\Omega resistor form the left branch, a 3 Ω3\,\Omega resistor the middle branch, and a 44 V battery with a 2 Ω2\,\Omega resistor the right branch. Both batteries have their positive terminal on top. The arrows show the directions CHOSEN for the currents I1I_1, I2I_2, I3I_3, in amperes.

9 V1 ΩI13 ΩI34 V2 ΩI2++
  • a) Write the junction equation at the top node, and the loop equations for the left loop and the right loop.
  • b) Write the augmented matrix and reduce it, naming every operation.
  • c) Give the three currents and check them in the two loop equations.
  • d) One current is negative. What does the sign mean for the circuit, and should the problem be redone with that arrow reversed?
  • e) Write the equation of the OUTER loop, through both batteries. Show that adding it to the system produces a row of zeros, and explain why.

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  • a) I1+I2−I3=0I_1 + I_2 - I_3 = 0, I1+3I3=9I_1 + 3I_3 = 9, 2I2+3I3=42I_2 + 3I_3 = 4
  • b) (11−100−1490012)\left(\begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 0 & -1 & 4 & 9 \\ 0 & 0 & 1 & 2 \end{array}\right)
  • c) I1=3I_1 = 3 A, I2=−1I_2 = -1 A, I3=2I_3 = 2 A
  • d) 11 A flows DOWN the right branch, into the ++ terminal: the 99 V battery charges the 44 V one. No need to redo.
  • e) I1−2I2=5I_1 - 2I_2 = 5, the left loop minus the right loop: it reduces to (0,0,0∣0)(0, 0, 0 \mid 0).
a)

At the top node, I1I_1 and I2I_2 arrive and I3I_3 leaves: I1+I2=I3I_1 + I_2 = I_3, that is I1+I2−I3=0I_1 + I_2 - I_3 = 0. Left loop, going up the left branch and down the middle one, in the directions of I1I_1 and I3I_3: the battery raises the potential by 99 V and the resistors drop 1⋅I11 \cdot I_1 and 3⋅I33 \cdot I_3, so I1+3I3=9I_1 + 3I_3 = 9. Right loop, going up the right branch and down the middle one, in the directions of I2I_2 and I3I_3: 2I2+3I3=42I_2 + 3I_3 = 4. The bottom node gives I3=I1+I2I_3 = I_1 + I_2 again, the same equation: a circuit with two nodes gives one useful junction equation, not two.

b)

The augmented matrix, unknowns in the order I1I_1, I2I_2, I3I_3, is (11−1010390234)\left(\begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 1 & 0 & 3 & 9 \\ 0 & 2 & 3 & 4 \end{array}\right). R2→R2−R1R_2 \to R_2 - R_1 gives (0,−1,4∣9)(0, -1, 4 \mid 9). R3→R3+2R2R_3 \to R_3 + 2R_2 gives (0,0,11∣22)(0, 0, 11 \mid 22), and R3→111R3R_3 \to \frac{1}{11}R_3 gives (0,0,1∣2)(0, 0, 1 \mid 2). The echelon form is (11−100−1490012)\left(\begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 0 & -1 & 4 & 9 \\ 0 & 0 & 1 & 2 \end{array}\right). Choosing R3+2R2R_3 + 2R_2 rather than dividing R2R_2 by −1-1 first is a matter of taste; what matters is that each step is named, because the marks of this question are in the method.

c)

Back substitution: I3=2I_3 = 2. The second row, −I2+4I3=9-I_2 + 4I_3 = 9, gives I2=8−9=−1I_2 = 8 - 9 = -1. The first, I1=I3−I2I_1 = I_3 - I_2, gives I1=2+1=3I_1 = 2 + 1 = 3. So I1=3I_1 = 3 A, I2=−1I_2 = -1 A, I3=2I_3 = 2 A. Check in the loop equations, not in the echelon form: left, 1(3)+3(2)=91(3) + 3(2) = 9; right, 2(−1)+3(2)=42(-1) + 3(2) = 4. The junction holds too: 3+(−1)=23 + (-1) = 2.

d)

The arrows were guesses, and the algebra corrects a wrong guess by a sign. I2=−1I_2 = -1 A means that a current of 11 A flows in the right branch in the direction OPPOSITE to the arrow, that is downwards, entering the 44 V battery by its positive terminal. The stronger 99 V battery is pushing current backwards through the weaker one: the 44 V battery is being charged. There is no need to redo the problem with the arrow reversed: the equations were written consistently with the arrows, so the answer is already correct, and reversing the arrow would only give +1+1 A for the same physical current. What must NOT be done is to drop the minus sign in the final answer while keeping the original arrow.

e)

Go around the outer loop clockwise: up the left branch, across the top, down the right branch. Up the left branch, the battery gives +9+9 V and the resistor drops 1⋅I11 \cdot I_1. Down the right branch, the battery is crossed from ++ to −-, a loss of 44 V, and the resistor is crossed AGAINST the arrow of I2I_2, so its drop counts as −2I2-2I_2. The equation is 9−4=I1−2I29 - 4 = I_1 - 2I_2, that is I1−2I2=5I_1 - 2I_2 = 5; check: 3+2=53 + 2 = 5. Add it as a fourth row (1,−2,0∣5)(1, -2, 0 \mid 5). R4→R4−R1R_4 \to R_4 - R_1 gives (0,−3,1∣5)(0, -3, 1 \mid 5), R4→R4−3R2R_4 \to R_4 - 3R_2 gives (0,0,−11∣−22)(0, 0, -11 \mid -22), and R4→R4+11R3R_4 \to R_4 + 11R_3 gives (0,0,0∣0)(0, 0, 0 \mid 0). The outer loop is the left loop minus the right loop, (I1+3I3)−(2I2+3I3)=9−4(I_1 + 3I_3) - (2I_2 + 3I_3) = 9 - 4, so it carries no new information, and a row that is a combination of the others always reduces to zeros. That is why only the inner loops are written: extra loops never hurt, they just cost time.

Exercise 10: Two more applications: balancing a reaction, mixing fertilizer

12 answers to check

Balancing a chemical equation and mixing products to hit a target are both linear systems, and both end with a question the algebra alone does not settle: which of the solutions is the one the situation asks for.

(1) Ammonia reduces copper(II) oxide: x1 NH3+x2 CuO→x3 N2+x4 Cu+x5 H2Ox_1\,\mathrm{NH_3} + x_2\,\mathrm{CuO} \to x_3\,\mathrm{N_2} + x_4\,\mathrm{Cu} + x_5\,\mathrm{H_2O}.

(2) A garden centre sells three fertilizers in bags. A bag of brand A holds 22 kg of nitrogen, 11 kg of phosphorus and 11 kg of potassium; a bag of brand B, 11, 22 and 33 kg; a bag of brand C, 11, 11 and 22 kg. A farmer needs exactly 1212 kg of nitrogen, 1111 kg of phosphorus and 1717 kg of potassium.

  • a) Write one equation per element for the reaction, N, H, Cu and O, and reduce the augmented matrix.
  • b) Give all the solutions, then the balanced equation with the smallest positive whole-number coefficients. Check it atom by atom.
  • c) Write the system for the numbers of bags aa, bb, cc of brands A, B, C and solve it by elimination.
  • d) Brand C is discontinued. Show by row reduction that the target can no longer be met exactly with brands A and B.

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  • a) Pivots on x1x_1 to x4x_4, x5x_5 free: x1=23x5x_1 = \frac{2}{3}x_5, x2=x5x_2 = x_5, x3=13x5x_3 = \frac{1}{3}x_5, x4=x5x_4 = x_5.
  • b) 2 NH3+3 CuO→N2+3 Cu+3 H2O2\,\mathrm{NH_3} + 3\,\mathrm{CuO} \to \mathrm{N_2} + 3\,\mathrm{Cu} + 3\,\mathrm{H_2O} (t=3t = 3)
  • c) a=3a = 3, b=2b = 2, c=4c = 4 bags
  • d) Last row (0,0∣8)(0, 0 \mid 8): inconsistent.
a)

Count each element on both sides. Nitrogen: x1=2x3x_1 = 2x_3. Hydrogen: 3x1=2x53x_1 = 2x_5. Copper: x2=x4x_2 = x_4. Oxygen: x2=x5x_2 = x_5. Every unknown goes to the left, so every right-hand side is 00: (10−20003000−20010−1000100−10)\left(\begin{array}{ccccc|c} 1 & 0 & -2 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & -2 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 \end{array}\right). R2→R2−3R1R_2 \to R_2 - 3R_1 gives (0,0,6,0,−2∣0)(0, 0, 6, 0, -2 \mid 0). R2↔R3R_2 \leftrightarrow R_3 puts the copper row second. R4→R4−R2R_4 \to R_4 - R_2 gives (0,0,0,1,−1∣0)(0, 0, 0, 1, -1 \mid 0). R3→16R3R_3 \to \frac{1}{6}R_3 gives (0,0,1,0,−13∣0)(0, 0, 1, 0, -\frac{1}{3} \mid 0). To reach the reduced form, R1→R1+2R3R_1 \to R_1 + 2R_3 gives (1,0,0,0,−23∣0)(1, 0, 0, 0, -\frac{2}{3} \mid 0) and R2→R2+R4R_2 \to R_2 + R_4 gives (0,1,0,0,−1∣0)(0, 1, 0, 0, -1 \mid 0). Pivots in the first four columns; the column of x5x_5 has none, so x5x_5 is free. A column of zeros on the right stays zero under every row operation, so it can be carried along without a thought.

b)

With x5=tx_5 = t: x1=23tx_1 = \frac{2}{3}t, x2=tx_2 = t, x3=13tx_3 = \frac{1}{3}t, x4=tx_4 = t, x5=tx_5 = t, for any real tt. Algebraically every value of tt works; chemically only positive whole numbers make sense, and the smallest tt that clears both thirds is t=3t = 3: x1=2x_1 = 2, x2=3x_2 = 3, x3=1x_3 = 1, x4=3x_4 = 3, x5=3x_5 = 3. The balanced equation is 2 NH3+3 CuO→N2+3 Cu+3 H2O2\,\mathrm{NH_3} + 3\,\mathrm{CuO} \to \mathrm{N_2} + 3\,\mathrm{Cu} + 3\,\mathrm{H_2O}. Atom count: 22 N on each side; 66 H on the left, 3×2=63 \times 2 = 6 on the right; 33 Cu and 33 O on each side. Two traps here: stopping at t=1t = 1 and handing in coefficients 23\frac{2}{3} and 13\frac{1}{3}, and taking t=0t = 0, which balances everything by writing no reaction at all.

c)

One equation per nutrient: nitrogen 2a+b+c=122a + b + c = 12, phosphorus a+2b+c=11a + 2b + c = 11, potassium a+3b+2c=17a + 3b + 2c = 17. R1↔R2R_1 \leftrightarrow R_2 puts (1,2,1∣11)(1, 2, 1 \mid 11) on top. R2→R2−2R1R_2 \to R_2 - 2R_1 gives (0,−3,−1∣−10)(0, -3, -1 \mid -10) and R3→R3−R1R_3 \to R_3 - R_1 gives (0,1,1∣6)(0, 1, 1 \mid 6). R2↔R3R_2 \leftrightarrow R_3 brings the convenient row up, and R3→R3+3R2R_3 \to R_3 + 3R_2 gives (0,0,2∣8)(0, 0, 2 \mid 8). So c=4c = 4, b=6−c=2b = 6 - c = 2, and a=11−2b−c=3a = 11 - 2b - c = 3. Check: nitrogen 6+2+4=126 + 2 + 4 = 12, phosphorus 3+4+4=113 + 4 + 4 = 11, potassium 3+6+8=173 + 6 + 8 = 17. The farmer buys 33 bags of A, 22 of B and 44 of C, and the answer is admissible: whole, non-negative numbers of bags.

d)

Without C the system is 2a+b=122a + b = 12, a+2b=11a + 2b = 11, a+3b=17a + 3b = 17: three equations, two unknowns. The same moves give: R1↔R2R_1 \leftrightarrow R_2, then R2→R2−2R1R_2 \to R_2 - 2R_1 gives (0,−3∣−10)(0, -3 \mid -10) and R3→R3−R1R_3 \to R_3 - R_1 gives (0,1∣6)(0, 1 \mid 6); R2↔R3R_2 \leftrightarrow R_3, then R3→R3+3R2R_3 \to R_3 + 3R_2 gives (0,0∣8)(0, 0 \mid 8), that is 0=80 = 8. The system is inconsistent: no combination of A and B hits the three targets at once. Any two targets can be met, nitrogen and phosphorus for instance with a=133a = \frac{13}{3} and b=103b = \frac{10}{3}, but then the potassium comes out at 133+10=433\frac{13}{3} + 10 = \frac{43}{3} kg, about 14.314.3 kg instead of 1717. The pivot in the augmented column is the algebra's way of saying that the third requirement is not a consequence of the first two.

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