MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: systems of linear equations and Gaussian elimination (MATH 133)

This sheet is not a summary of the chapter: you already have Nicholson and the lecture notes. It answers one question only, what makes students lose marks on systems of linear equations and Gaussian elimination in MATH 133 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam, where there is no calculator. The points are given to the method: a solution that names every row operation, $R_3 \to R_3 - 2R_1$ , earns marks even with a slip in the arithmetic, and a correct final answer with no visible reduction earns almost none.

The thread of the chapter

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 $k - 2$ before knowing that $k \neq 2$ , 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.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Gaussian elimination
Gauss-Jordan elimination
Augmented matrix
Elementary row operations
Row echelon form
Reduced row echelon form
Leading and free variables
Parametric general solution
Inconsistent system
Linear system with a parameter
Polynomial interpolation
Kirchhoff's laws

The essentials

The three legal moves, and the rule behind them

• Swap two rows: $R_i \leftrightarrow R_j$ .
• Multiply a row by a NON-ZERO number: $R_i \to cR_i$ with $c \neq 0$ .
• Add a multiple of ANOTHER row: $R_i \to R_i + cR_j$ with $j \neq i$ , and any $c$ , even an expression in $k$ , even $0$ .
• The rule behind all three: the row being replaced appears in its replacement with a non-zero coefficient. Then the step can be undone, and the system has exactly the same solutions before and after.
• Combined moves are fine when they respect the rule: $R_2 \to 3R_2 - 2R_1$ is legal, $R_2 \to R_3 - R_1$ is not.

The move

R_2 \to R_2 - R_1

replaces the line

x - y = -2

y = 3

, but the crossing point

(1, 3)

stays put: the lines change, the solution does not.

Name every operation on your paper, $R_3 \to R_3 - 2R_1$ and not an arrow with nothing on it: in MATH 133 the marks go to the method, and a marker cannot give them to a matrix that changes without explanation.

Echelon, reduced echelon, and what each one tells you

• Row echelon form: zero rows at the bottom, each leading entry strictly to the right of the leading entry above it. Enough to read the verdict and back-substitute.
• Reduced row echelon form: in addition, every leading entry is $1$ and is the only non-zero entry of its column. It shows the solution directly.
• A matrix has MANY echelon forms and exactly ONE reduced form. All its echelon forms share the POSITIONS of the pivots, hence their number.
• Pivot variable: its column holds a leading entry. Free variable: its column holds none. Where the column sits in the matrix decides nothing.
• Inconsistent: a row $(0 \; 0 \; \cdots \; 0 \mid c)$ with $c \neq 0$ . Nothing else makes a system inconsistent.

When your echelon form differs from the answer key, do not panic: reduce both to the reduced form, or compare the pivot columns and the final solution. Two different echelon forms can both be right.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Copying the equations into the matrix without lining up the variables

the whole question: every later step is exact arithmetic on the wrong system

What not to write

“For $2y - x + 3z = 7$ and $z = 5 - 2x$ , the rows are $(2, -1, 3 \mid 7)$ and $(1, -2 \mid 5)$ .”

What to write

“In the order $x, y, z$ : $-x + 2y + 3z = 7$ gives $(-1, 2, 3 \mid 7)$ , and $2x + 0y + z = 5$ gives $(2, 0, 1 \mid 5)$ .”

Why: A column is a VARIABLE, not a position in the sentence. Rewrite each equation as coefficients of $x$ , $y$ , $z$ in that order equals a constant before copying anything, write $0$ for a missing variable, and change the sign of every term that crosses the equals sign.

2. Replacing a row by a combination that no longer contains it

the whole question, and the final check may not see it

What not to write

“ $R_2 \to R_3 - R_1$ , to get a zero in the corner faster.”

What to write

“ $R_3 \to R_3 - R_1$ : the row being replaced keeps its coefficient $1$ , so the step can be undone.”

Why: The old second equation is deleted, and a system with one equation fewer usually has MORE solutions. With $-x + 2y + 3z = 7$ , $2x + z = 5$ , $3x - y - 2z = 1$ , the damaged system accepts $(3, -4, 6)$ , which gives $2x + z = 12$ . The true solution $(2, 3, 1)$ survives, which is why checking only the expected answer shows nothing.

3. Doing two operations at once on the old rows

2 marks, and a system that quietly gains infinitely many solutions

What not to write

“ $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_1$ , in one step.”

What to write

“ $R_1 \to R_1 - R_2$ first; then, on the NEW matrix, $R_2 \to R_2 - R_1$ .”

Why: Done at the same time on $x + y = 3$ and $x - y = 1$ , the two new rows are $(0, 2 \mid 2)$ and $(0, -2 \mid -2)$ , opposites of each other: an equation has vanished and $(5, 1)$ now passes although $5 + 1 \neq 3$ . Done one after the other, each step can be undone and nothing is lost.

4. Dividing by an expression in $k$ before splitting the cases

the whole case $k = 2$, typically a third of the question

What not to write

“The last row is $(k-2)(k+2)z = k - 2$ , so $(k+2)z = 1$ and $z = \frac{1}{k+2}$ : one solution for every $k \neq -2$ .”

What to write

“If $k \neq \pm 2$ , $z = \frac{1}{k+2}$ and the solution is unique. If $k = -2$ , the row reads $0 = -4$ : no solution. If $k = 2$ , it reads $0 = 0$ : $z$ is free, infinitely many solutions.”

Why: Dividing by $k - 2$ is multiplying by $\frac{1}{k - 2}$ , which does not exist at $k = 2$ : the step is illegal for exactly the value that matters. The student still finds ONE solution at $k = 2$ , $z = \frac{1}{4}$ , and it really is a solution, which is why the error survives substitution: the number is right, the word “exactly” is wrong.

5. Scaling the row being replaced by an expression that can vanish

2 marks, and a false case at $k = 1$

What not to write

“ $R_3 \to (k-1)R_3 - R_2$ , to clear the entry below the pivot.”

What to write

“ $R_3 \to R_3 - (k-1)R_2$ : the expression multiplies the OTHER row, so the step is legal for every $k$ , $k = 1$ included.”

Why: At $k = 1$ the faulty move replaces $R_3$ by $-R_2$ : the third equation is destroyed and the system gains solutions that do not exist. The multiplier of the other row may be anything, even zero; the coefficient of the row being replaced may never be zero. That is the whole difference between the two lines.

6. Reading a zero row as infinitely many solutions

2 marks, and a parameter that does not exist in the answer

What not to write

“After reduction the last row is $(0, 0 \mid 0)$ , so there are infinitely many solutions.”

What to write

“Both variable columns hold a pivot, so no variable is free: the solution is unique, $(x, y) = (2, 1)$ . The zero row only says that the third equation follows from the first two.”

Three equations, two unknowns, one zero row after reduction, and still ONE solution: the three lines meet at the same point

(2, 1)

Why: Infinitely many solutions come from a COLUMN without a pivot in a consistent system, never from a row without one. Three lines through one point, as in the figure, give a zero row and a single solution.

7. Declaring infinitely many solutions before testing consistency

the whole question

What not to write

“In $\left(\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right)$ the column of $y$ has no pivot, so $y$ is free and there are infinitely many solutions.”

What to write

“The second row reads $0 = 1$ : the system is inconsistent, and the free column is irrelevant.”

Why: The verdict has an ORDER. First look for a row $(0 \; \cdots \; 0 \mid c)$ with $c \neq 0$ ; only if there is none do the free columns count. A free variable in an inconsistent system is free to take any value in an empty set.

8. Using a pivot variable as a parameter, or one parameter too many

2 to 3 marks: the set written contains points that are not solutions

What not to write

“From $\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 3 \\ 0 & 0 & 1 & 2 & -1 \end{array}\right)$ : $x_1 = 3 - 2r + t$ , $x_2 = r$ , $x_3 = s$ , $x_4 = t$ .”

What to write

“ $x_2 = s$ and $x_4 = t$ are free and $x_3 = -1 - 2t$ is forced: $(x_1, x_2, x_3, x_4) = (3 - 2s + t, s, -1 - 2t, t)$ , $s, t \in \mathbb{R}$ .”

Why: One parameter per column WITHOUT a pivot, no more. With three parameters, $r = s = t = 0$ gives $(3, 0, 0, 0)$ , which fails the second row, $x_3 + 2x_4 = -1$ . Count the non-pivot columns, here $4 - 2 = 2$ , before writing a single letter.

Which method to choose

What the echelon form says, read in this order

Reduce to an echelon form first, then ask these questions in THIS order

Three echelon forms, three verdicts: a pivot after the bar kills the system, a pivot in every variable column gives one point, a column without a pivot gives a family.

If a row $(0 \; \cdots \; 0 \mid c)$ with $c \neq 0$ → no solution, stop reading

Example: $(0, 0, 0 \mid -4)$ at $k = -2$
If no such row, and every variable column holds a pivot → exactly one solution, found by back substitution

Example: $(0, 0, 7 \mid 7)$ gives $z = 1$ , then $y = 3$ and $x = 2$
If no such row, and some variable columns hold no pivot → infinitely many solutions, one parameter per free column

Example: pivots in columns $1$ and $3$ of $4$ : $x_2 = s$ , $x_4 = t$
If a pivot that depends on a parameter $k$ → factor it, list the values where it vanishes, and study each one SEPARATELY, right-hand side included

Example: $(k - 2)(k + 2)$ : study $k = 2$ , then $k = -2$ , then all other $k$

Zero rows are absent from this tree on purpose: a row $(0 \; \cdots \; 0 \mid 0)$ decides nothing, it only records an equation that followed from the others.

What the unknowns are, in each application

Decide what the unknowns are before writing a single equation

If a curve through given points → unknowns = the COEFFICIENTS of the polynomial, one equation per point

Example: $y = a + bx + cx^2$ through $(1, 2)$ gives $a + b + c = 2$
If a chemical equation to balance → unknowns = the coefficients, one equation per ELEMENT, then choose the parameter that clears the fractions

Example: $x_1 = \frac{2}{3}t$ , $x_3 = \frac{1}{3}t$ : take $t = 3$
If an electrical circuit → unknowns = the branch currents; junction equations for all nodes but one, a loop equation per inner loop

Example: $I_1 + I_2 = I_3$ , $I_1 + 3I_3 = 9$ , $2I_2 + 3I_3 = 4$
If products mixed to hit a target → unknowns = the quantity of each product, one equation per ingredient

Example: $2a + b + c = 12$ kg of nitrogen

The algebra gives every solution; the situation then chooses: whole numbers for a chemical equation, a direction for a negative current, non-negative quantities for a mix.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Discussing a system with a parameter $k$

When to use it: The statement says “for which values of $k$ ” or “discuss according to $k$ ”

1 Reduce with numbers, or with expressions in $k$ that multiply the OTHER row only. Never scale the row being replaced by an expression in $k$ , never divide by one.
2 Factor the last pivot and list the values of $k$ where it vanishes. Write them down before anything else: they are the only candidates for a special case.
3 For $k$ outside that list, say that every variable column holds a pivot, so the solution is unique; only NOW may you divide.
4 For each listed value, substitute it into the echelon form and read the last row: $(0 \; \cdots \; 0 \mid c)$ with $c \neq 0$ means no solution, $(0 \; \cdots \; 0 \mid 0)$ means look for the free columns.
5 Conclude with one sentence per case, and solve the infinite case in parametric form if it is asked.

Concluding sentence

“If $k \neq 2$ and $k \neq -2$ , the system has exactly one solution; if $k = -2$ it has none, since the last row reads $0 = -4$ ; if $k = 2$ it has infinitely many, with $z$ free.”

The trap: Studying a special value in a row that has already been divided by $k - 2$ : the case has disappeared before you look for it.

Marking: Typically 3 marks for a reduction with named operations, 2 for the list of special values, 1 per case correctly concluded, and 2 for the parametric solution of the infinite case.

Writing the general solution

When to use it: The echelon form is consistent and at least one variable column has no pivot

1 Take the matrix to the REDUCED form: each pivot variable then appears in exactly one equation.
2 Name the free variables by their columns and give each a parameter: $x_2 = s$ , $x_4 = t$ .
3 Solve each row for its pivot variable in terms of the parameters.
4 Write the solution as a list, or as one vector plus multiples of fixed vectors, and check the fixed vector, all parameters at $0$ , in the ORIGINAL system.

Concluding sentence

“The general solution is $(x_1, x_2, x_3, x_4) = (3, 0, -1, 0) + s(-2, 1, 0, 0) + t(1, 0, -2, 1)$ , with $s$ and $t$ any real numbers.”

The trap: Forgetting “with $s$ and $t$ any real numbers”: without it the letters are unknowns, not parameters, and the line describes nothing.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Substitute into the ORIGINAL system

The echelon form only checks your back substitution; the original system also catches a bad row operation. Substitute every equation, not just the first.

$(2, 3, 1)$ in $2y - x + 3z = 7$ : $6 - 2 + 3 = 7$ . Checked only in the echelon form, an illegal $R_2 \to R_3 - R_1$ would go unnoticed.
Count the parameters

Number of parameters = number of unknowns minus number of pivots. Any other count is wrong before you look at a single entry.

$4$ unknowns, $2$ pivots: exactly $2$ parameters. Three letters on the page means one of them is a pivot variable in disguise.
Test one value of k in each case

Pick one value of $k$ in each case and solve the numerical system quickly. Each must land in the case you announced.

$k = 1$ must give exactly one solution, $(2, -\frac{1}{3}, \frac{1}{3})$ . If it gives a zero row, the factorisation of the pivot is wrong.
Check the context

A balanced reaction has whole coefficients and the same atoms on both sides; the currents of a circuit also satisfy the outer loop.

$2\,\mathrm{NH_3} + 3\,\mathrm{CuO} \to \mathrm{N_2} + 3\,\mathrm{Cu} + 3\,\mathrm{H_2O}$ : $2$ N, $6$ H, $3$ Cu, $3$ O on each side.

The typical problem, taken apart

A parameter question where the three cases are not all there

Discuss, according to the real parameter $k$ , the number of solutions of $\begin{cases} x + y + z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2 \end{cases}$ , then solve it for $k = 1$ .

Step 1

$R_2 \to R_2 - 2R_1$ gives $(0, 1, k - 2 \mid 1)$ and $R_3 \to R_3 - R_1$ gives $(0, k - 1, 2 \mid 1)$ .

Why

Both multipliers are numbers, so both moves are legal for every $k$ at once. The first pivot is a plain $1$ , which is why that row stays on top.

Step 2

$R_3 \to R_3 - (k - 1)R_2$ gives $(0, 0, 2 - (k - 1)(k - 2) \mid 1 - (k - 1))$ , that is $(0, 0, -k^2 + 3k \mid 2 - k)$ .

Why

The expression $k - 1$ multiplies the OTHER row, so the move is legal even at $k = 1$ , where it subtracts nothing. Resist dividing $R_3$ by $k - 1$ to make a leading $1$ : that move is illegal at $k = 1$ , and it is not needed.

Step 3

Factor the last pivot: $-k^2 + 3k = k(3 - k)$ , which vanishes at $k = 0$ and $k = 3$ only.

Why

These two values are the only candidates for a special case. Listing them BEFORE solving is what the marker looks for, and it stops you from dividing by zero two lines later.

Step 4

At $k = 0$ the last row is $(0, 0, 0 \mid 2)$ , and at $k = 3$ it is $(0, 0, 0 \mid -1)$ : no solution in either case. For every other $k$ , the three columns hold a pivot: exactly one solution.

Why

There is no value with infinitely many solutions here, because the right-hand side is non-zero at both special values. The three cases are possibilities, not a quota, and inventing an infinite case costs as much as missing one.

Step 5

For $k = 1$ the last row is $(0, 0, 2 \mid 1)$ , so $z = \frac{1}{2}$ ; then $y + (1 - 2)z = 1$ gives $y = \frac{3}{2}$ , and $x = 1 - y - z = -1$ .

Why

With $k$ fixed, the pivot $k(3 - k) = 2$ is a number, and dividing by it is now legal. The division was never wrong in itself; it was wrong before the cases were split.

Step 6

Check in the original system at $k = 1$ : $-1 + \frac{3}{2} + \frac{1}{2} = 1$ , $-2 + \frac{9}{2} + \frac{1}{2} = 3$ , $-1 + \frac{3}{2} + \frac{3}{2} = 2$ .

Why

The ORIGINAL equations, not the echelon form: this is the only check that also catches a wrong row operation, and it takes thirty seconds.

The conclusion, written out

“If $k = 0$ or $k = 3$ the system has no solution; for every other real $k$ it has exactly one; for $k = 1$ that solution is $(x, y, z) = (-1, \frac{3}{2}, \frac{1}{2})$ .”

The classic mistake on this problem: Announcing “infinitely many solutions for $k = 3$ ” because the coefficient part of the last row vanishes, without reading its right-hand side, which is $-1$ .

Learn by heart

• Three legal moves: swap, scale by a NON-ZERO number, add a multiple of ANOTHER row.
• The row being replaced must appear in its replacement with a non-zero coefficient.
• Echelon forms are many, the reduced row echelon form is UNIQUE, and the pivot columns are the same in all of them.
• Inconsistent exactly when a row reads $(0 \; \cdots \; 0 \mid c)$ with $c \neq 0$ . Test it FIRST.
• Consistent and a column without a pivot: infinitely many solutions, one parameter per free column.
• With $k$ : never divide by an expression in $k$ ; factor the last pivot, then study each root separately, right-hand side included.
• A zero row decides nothing: it is a redundant equation, not a free variable.
• Always check in the ORIGINAL system.

Frequently asked questions

How do I know if a system has no solution, one solution or infinitely many?

Reduce the augmented matrix to an echelon form, then read it in order. If a row has zeros everywhere except in the last column, there is no solution. Otherwise, if every variable column holds a pivot, there is exactly one. Otherwise some variable is free and there are infinitely many, one parameter for each column without a pivot.

Is the row echelon form of a matrix unique?

No. Different sequences of row operations give different echelon forms, with different entries and even different pivot values. What every echelon form of a matrix shares is the position of its pivots. The reduced row echelon form, on the other hand, is unique: two correct reductions always end on exactly the same reduced matrix, which makes it the right thing to compare with an answer key.

Why can't I divide by k minus 2 when solving a system with a parameter?

Because dividing by k minus 2 means multiplying by one over k minus 2, which does not exist when k equals 2. The operation is illegal for exactly the value that needs studying, and after it that case has silently disappeared. Factor the pivot, list the values where it vanishes, study each of them on its own, and divide only in the general case.

What is the difference between a pivot variable and a free variable?

A pivot variable has a leading entry in its column of the echelon form; a free variable has none. Free variables can take any value and become the parameters of the general solution, while pivot variables are then forced by the equations. Being free depends only on the absence of a pivot, never on whether the variable comes last.

Do I need a calculator for Gaussian elimination in MATH 133?

No, and you will not have one on the exam. The systems are chosen so that every pivot reduces by hand to whole numbers or simple fractions. Pick moves that keep integers, such as swapping a row with a leading one to the top, and name every operation you perform, because the marks are given to the method as much as to the final answer.

Practise it

Corrected exercises: Systems of linear equations and Gaussian elimination, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Next sheet Homogeneous systems and the rank theorem

Looking for a MATH 133 tutor in Montreal?

Get in touch for a first session. The first chapter of linear algebra is where the method is built: legal moves, one verdict read off the echelon form, and every case of a parameter accounted for.

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Revision sheet: systems of linear equations and Gaussian elimination (MATH 133)

The essentials

The three legal moves, and the rule behind them

Echelon, reduced echelon, and what each one tells you

The mistakes that cost marks

1. Copying the equations into the matrix without lining up the variables

2. Replacing a row by a combination that no longer contains it

3. Doing two operations at once on the old rows

4. Dividing by an expression in kkk before splitting the cases

5. Scaling the row being replaced by an expression that can vanish

6. Reading a zero row as infinitely many solutions

7. Declaring infinitely many solutions before testing consistency

8. Using a pivot variable as a parameter, or one parameter too many

Which method to choose

What the echelon form says, read in this order

What the unknowns are, in each application

How the answer is expected to be written

Discussing a system with a parameter kkk

Writing the general solution

Check before you hand in

The typical problem, taken apart

A parameter question where the three cases are not all there

Learn by heart

Frequently asked questions

How do I know if a system has no solution, one solution or infinitely many?

Is the row echelon form of a matrix unique?

Why can't I divide by k minus 2 when solving a system with a parameter?

What is the difference between a pivot variable and a free variable?

Do I need a calculator for Gaussian elimination in MATH 133?

Corrected exercises: Systems of linear equations and Gaussian elimination, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?

4. Dividing by an expression in $k$ before splitting the cases

Discussing a system with a parameter $k$