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

a) Every equation is rewritten as (coefficient of $x$) $+$ (coefficient of $y$) $+$ (coefficient of $z$) $=$ constant BEFORE a single entry is copied. The first equation reads $-x + 2y + 3z = 7$: its terms were out of order, and copying them as written would have put the $2$ in the $x$ column. The second, $z = 5 - 2x$, becomes $2x + 0y + z = 5$: the variable $y$ is absent, and the absence is written as a $0$, never skipped, otherwise every entry to its right slides one column to the left. The third, $3x - y = 2z + 1$, becomes $3x - y - 2z = 1$: the $2z$ changes sign as it crosses the equals sign. The augmented matrix is $\left(\begin{array}{ccc|c} -1 & 2 & 3 & 7 \\ 2 & 0 & 1 & 5 \\ 3 & -1 & -2 & 1 \end{array}\right)$. 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 $R_2$ by $3$, then adds $-2R_3$ to it, two elementary operations in a row. The row being REPLACED keeps a non-zero coefficient, $3$, so the step is undone by $R_2 \to \frac{1}{3}(R_2 + 2R_3)$. (iv) is the third elementary operation, adding a multiple of another row, undone by $R_3 \to R_3 - 3R_1$: legal. (iii) is ILLEGAL. The new second row $R_3 - R_1$ does not contain $R_2$ at all, so the information carried by the equation $2x + z = 5$ 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 $0$.

c) Make the leading entry of $R_1$ equal to $1$: $R_1 \to -R_1$ gives $(1, -2, -3 \mid -7)$. Clear the first column: $R_2 \to R_2 - 2R_1$ gives $(0, 4, 7 \mid 19)$ and $R_3 \to R_3 - 3R_1$ gives $(0, 5, 7 \mid 22)$. Rather than dividing by $4$ and meeting quarters, use $R_3 \to R_3 - R_2$, which gives $(0, 1, 0 \mid 3)$, then $R_2 \leftrightarrow R_3$ to bring that convenient row up. Now $R_3 \to R_3 - 4R_2$ gives $(0, 0, 7 \mid 7)$ and $R_3 \to \frac{1}{7}R_3$ gives $(0, 0, 1 \mid 1)$. The echelon form is $\left(\begin{array}{ccc|c} 1 & -2 & -3 & -7 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}\right)$. Back substitution: $z = 1$, $y = 3$, and $x - 2y - 3z = -7$ gives $x = -7 + 6 + 3 = 2$. Check in the ORIGINAL equations: $2(3) - 2 + 3(1) = 7$, $1 = 5 - 2(2)$, $3(2) - 3 = 2(1) + 1$. All three hold, so $(x, y, z) = (2, 3, 1)$. 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 $R_1$, $R_3 - R_1$ and $R_3$, that is $-x + 2y + 3z = 7$, $4x - 3y - 5z = -6$ and $3x - y - 2z = 1$. For $(3, -4, 6)$: $-3 - 8 + 18 = 7$, $12 + 12 - 30 = -6$ and $9 + 4 - 12 = 1$, so it satisfies all three. But in the original second equation, $2x + z = 6 + 6 = 12 \neq 5$: 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 $(2, 3, 1)$ and $(3, -4, 6)$. Notice the trap inside the trap: the true solution $(2, 3, 1)$ 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 $x + y = 4$ and $x - y = -2$, so it also satisfies their difference, $(x - y) - (x + y) = -2 - 4$, that is $-2y = -6$, or $y = 3$. The crossing point $(1, 3)$ therefore lies on the new line, and the argument works whatever multiple of $R_1$ is subtracted: every operation of the third kind swings the line of $R_2$ around the solution point like a door on its hinge. No new crossing point can appear because the move can be undone: $R_2 \to R_2 + R_1$ brings the old line back, so a point lying on $R_1$ 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.

a) $R_1 \to \frac{1}{2}R_1$ gives $(1, 2, -1 \mid 1)$. Then $R_2 \to R_2 - R_1$ gives $(0, 1, 2 \mid 5)$ and $R_3 \to R_3 - 3R_1$ gives $(0, -1, -1 \mid -3)$. Finally $R_3 \to R_3 + R_2$ gives $(0, 0, 1 \mid 2)$. Her echelon form is $\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 2 \end{array}\right)$. Dividing the first row by $2$ 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) $R_1 \leftrightarrow R_2$ puts $(1, 3, 1 \mid 6)$ on top, with a leading $1$ for free. Then $R_2 \to R_2 - 2R_1$ gives $(2, 4, -2 \mid 2) - (2, 6, 2 \mid 12) = (0, -2, -4 \mid -10)$ and $R_3 \to R_3 - 3R_1$ gives $(3, 5, -4 \mid 0) - (3, 9, 3 \mid 18) = (0, -4, -7 \mid -18)$. Finally $R_3 \to R_3 - 2R_2$ gives $(0, 0, 1 \mid 2)$. His echelon form is $\left(\begin{array}{ccc|c} 1 & 3 & 1 & 6 \\ 0 & -2 & -4 & -10 \\ 0 & 0 & 1 & 2 \end{array}\right)$. 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 $1$.

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, $1$ for A and $-2$ 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 $1$, $2$ and $3$, 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: $R_2 \to R_2 - 2R_3$ gives $(0, 1, 0 \mid 1)$, $R_1 \to R_1 + R_3$ gives $(1, 2, 0 \mid 3)$, then $R_1 \to R_1 - 2R_2$ gives $(1, 0, 0 \mid 1)$. From B: $R_2 \to -\frac{1}{2}R_2$ gives $(0, 1, 2 \mid 5)$, $R_2 \to R_2 - 2R_3$ gives $(0, 1, 0 \mid 1)$, $R_1 \to R_1 - R_3$ gives $(1, 3, 0 \mid 4)$, then $R_1 \to R_1 - 3R_2$ gives $(1, 0, 0 \mid 1)$. Both routes end on $\left(\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right)$, 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: $(x, y, z) = (1, 1, 2)$. Check in the original system: $2 + 4 - 4 = 2$, $1 + 3 + 2 = 6$, $3 + 5 - 8 = 0$. 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.

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

b) The pivots sit in columns $1$ and $3$: $x_1$ and $x_3$ are the pivot variables. Columns $2$ and $4$ hold no pivot: $x_2$ and $x_4$ are free. Yes, $x_2$ 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 $1$ to column $3$ is exactly what created a free variable in the middle. A frequent wrong answer declares only $x_4$ 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, $x_2 = s$ and $x_4 = t$, then solve each row of the REDUCED form for its pivot variable: $x_1 = 3 - 2s + t$ and $x_3 = -1 - 2t$. The general solution is $(x_1, x_2, x_3, x_4) = (3 - 2s + t, s, -1 - 2t, t)$, with $s$ and $t$ any real numbers, or in vector form $(3, 0, -1, 0) + s(-2, 1, 0, 0) + t(1, 0, -2, 1)$. Check the fixed vector in the original system: $3 + 0 - 0 = 3$, $6 + 0 - 1 = 5$, $9 + 0 - 1 - 0 = 8$. The words “with $s$ and $t$ any real numbers” are part of the answer: without them the letters are unknowns and the line says nothing.

d) $x_2 = 0$ means $s = 0$. Then $x_1 = 3 + t = 0$ forces $t = -3$, so $x_4 = -3$ and $x_3 = -1 - 2(-3) = 5$. The solution is $(0, 0, 5, -3)$. Check in the original system: $0 + 0 - (-3) = 3$, $0 + 0 + 5 = 5$, $0 + 0 + 5 - (-3) = 8$. 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 $s$ and $t$.

e) Ana is right. Her fixed vector $(5, -1, -1, 0)$ is a solution: $5 - 2 - 0 = 3$, $10 - 4 - 1 = 5$, $15 - 6 - 1 - 0 = 8$; it is ours with $s = -1$. Her two direction vectors are ours, so as $s$ and $t$ 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 $x_3$ as free although its column holds a pivot, so $x_3$ is not free, it is forced to equal $-1 - 2x_4$. His list contains points that are not solutions: $r = s = t = 0$ gives $(3, 0, 0, 0)$, and the second equation gives $6 + 0 + 0 = 6 \neq 5$. Count before writing: the number of parameters equals the number of unknowns minus the number of pivots, here $4 - 2 = 2$.

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

b) Factor the last pivot: $k^2 - 4 = (k - 2)(k + 2)$. It vanishes for $k = 2$ and $k = -2$ only, so those are the only values that can be special, and each is studied separately in the echelon form. If $k \neq 2$ and $k \neq -2$, the three columns hold a pivot, no variable is free and no row is contradictory: exactly one solution. If $k = -2$, the last row becomes $(0, 0, 0 \mid -4)$, that is $0 = -4$: no solution. If $k = 2$, it becomes $(0, 0, 0 \mid 0)$: the system is consistent and column $3$ has lost its pivot, so $z$ 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 $k - 2$ is what separates $k = 2$ from $k = -2$, although both kill the same pivot.

c) Dividing by $k - 2$ is multiplying by $\frac{1}{k - 2}$, which does not exist when $k = 2$. The step is illegal for exactly the value that matters, and after it the case $k = 2$ has silently disappeared: his last row $(k + 2)z = 1$ still has a non-zero pivot at $k = 2$, so his work claims ONE solution there, $z = \frac{1}{4}$, $y = -1 + 2z = -\frac{1}{2}$ and $x = 2 - y - z = \frac{9}{4}$. The dangerous part is that this triple really is a solution for $k = 2$, it is the case $t = \frac{1}{4}$ 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 $k = 2$ 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 $k = 2$ the echelon form is $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right)$. Column $3$ has no pivot, so $z = t$ with $t$ any real number. The second row gives $y = -1 + 2t$, the first $x = 2 - y - z = 2 + 1 - 2t - t = 3 - 3t$. The solutions are $(x, y, z) = (3, -1, 0) + t(-3, 2, 1)$, a line in space. Check the third ORIGINAL equation at $k = 2$, where it reads $2x + 3y + 0z = 3$: $2(3 - 3t) + 3(-1 + 2t) = 6 - 6t - 3 + 6t = 3$ for every $t$, as it must.

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

a) $R_1 \to \frac{1}{k}R_1$ multiplies a row by $\frac{1}{k}$, which does not exist when $k = 0$. The step would force a separate case $k = 0$ before the reduction even starts, and a student who forgets that case has used an illegal operation. The fix costs nothing: $R_1 \leftrightarrow R_3$ brings $(1, 1, k \mid 1)$ to the top, whose leading entry is $1$ for every value of $k$. 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 $(1, 1, k \mid 1)$, $(1, k, 1 \mid 1)$ and $(k, 1, 1 \mid 1)$. $R_2 \to R_2 - R_1$ gives $(0, k - 1, 1 - k \mid 0)$. $R_3 \to R_3 - kR_1$ gives $(0, 1 - k, 1 - k^2 \mid 1 - k)$: this move is legal for every $k$, even $k = 0$, because the expression $k$ multiplies the OTHER row and the row being replaced keeps its coefficient $1$. Then $R_3 \to R_3 + R_2$ gives $(0, 0, 2 - k - k^2 \mid 1 - k)$. Factor: $2 - k - k^2 = (1 - k)(k + 2)$. The echelon form is $\left(\begin{array}{ccc|c} 1 & 1 & k & 1 \\ 0 & k - 1 & 1 - k & 0 \\ 0 & 0 & (1 - k)(k + 2) & 1 - k \end{array}\right)$. Two leading entries depend on $k$, $k - 1$ and $(1 - k)(k + 2)$, and none of them has been divided by.

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

d) For $k = 1$, put $y = s$ and $z = t$: the general solution is $(x, y, z) = (1 - s - t, s, t)$, with $s, t$ any real numbers. The student's triple is a solution, $\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$; it is the case $s = t = \frac{1}{3}$. It is NOT the only one: $(1, 0, 0)$, $(0, 1, 0)$ and $(2, -1, 0)$ are others. His division by $k - 1$ turned the second row into $y - z = 0$ and the third into $(k + 2)z = 1$, two equations that still have pivots at $k = 1$, 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 $k \neq 1, -2$, dividing is now legal. The third row gives $z = \frac{1 - k}{(1 - k)(k + 2)} = \frac{1}{k + 2}$. The second, $(k - 1)(y - z) = 0$ with $k - 1 \neq 0$, gives $y = z$. The first gives $x = 1 - y - kz = 1 - \frac{1 + k}{k + 2} = \frac{1}{k + 2}$. So $x = y = z = \frac{1}{k + 2}$, which the symmetry of the system predicted: exchanging the unknowns exchanges the equations. For $k = 0$, $x = y = z = \frac{1}{2}$; check: $y + z = 1$, $x + z = 1$, $x + y = 1$. The value $k = 0$, 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 $k$: it creates a fake case ($k = 0$) and hides a real one ($k = 1$).

a) Start from $\left(\begin{array}{cc|c} 1 & 2 & 4 \\ 1 & -1 & 1 \\ k & 1 & 5 \end{array}\right)$. $R_2 \to R_2 - R_1$ gives $(0, -3 \mid -3)$, and $R_2 \to -\frac{1}{3}R_2$ gives $(0, 1 \mid 1)$. $R_3 \to R_3 - kR_1$ gives $(0, 1 - 2k \mid 5 - 4k)$, legal for every $k$ since the expression multiplies the other row. Then $R_3 \to R_3 - (1 - 2k)R_2$ gives $(0, 0 \mid 5 - 4k - (1 - 2k)) = (0, 0 \mid 4 - 2k)$, 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: $k$ 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 $0 = 4 - 2k$. It is a contradiction unless $4 - 2k = 0$, that is $k = 2$. For $k = 2$ the rows give $y = 1$, then $x = 4 - 2y = 2$: the solution is $(2, 1)$. Check the third equation: $2(2) + 1 = 5$. In the picture, $k = 2$ 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 $k = 2$ right and the only interesting one wrong: counting equations is never the test, the last row is.

c) For $k = 1$ the third line is $x + y = 5$. First and second: subtracting gives $3y = 3$, so $(2, 1)$. First and third: subtracting gives $y = -1$, then $x = 6$, so $(6, -1)$. Second and third: adding gives $2x = 6$, so $x = 3$, $y = 2$, point $(3, 2)$. 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 $k = 1$ the last row reads $0 = 2$.

d) The third line $kx + y = 5$ has slope $-k$. The first, $x + 2y = 4$, has slope $-\frac{1}{2}$, and the second, $x - y = 1$, has slope $1$. So the third line is parallel to the first for $k = \frac{1}{2}$, where it reads $x + 2y = 10$, and to the second for $k = -1$, where it reads $x - y = -5$. In both cases the lines are parallel and DISTINCT, so they never meet, and the system has no solution: the last row reads $0 = 3$ and $0 = 6$. 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 $k$ is, the echelon form keeps a pivot in both variable columns, $1$ and $1$, 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 $-\frac{1}{2}$ and $1$, are different lines before $k$ enters the problem at all. So for this system the three textbook cases shrink to two: one solution for $k = 2$, none otherwise. A parameter question does not promise all three cases, and inventing one to fill the pattern costs as much as missing one.