MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: determinants, inverses and Cramer's rule (MATH 133)

This sheet is not a summary of section 3.2 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on the rules of the determinant, the adjugate and Cramer's rule in MATH 133 at McGill University, and which precise gesture avoids each loss.

Everything below comes back to one fact: $\det(AB) = \det A \, \det B$ . The inverse, the transpose, the power and the scalar rules are that fact read in four directions, and the adjugate and Cramer's rule are the place where $\det A$ ends up in a denominator. Every number on this sheet reduces by hand, as it must on an exam without a calculator.

Updated on October 17, 2026

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The thread of the chapter

The determinant is a MULTIPLICATIVE volume factor: products multiply, inverses divide, the transpose changes nothing, and a scalar in front of an $n \times n$ matrix scales $n$ edges, hence $k^{n}$ . The same number $\det A$ sits in the denominator of every entry of $A^{-1}$ and of every Cramer unknown, so $\det A = 0$ is where all of them stop, without ever saying that a system has no solution.

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

What is on this sheet

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

Topics covered

Multiplicativity of the determinant
Determinant of a transpose
Determinant of an inverse
Determinant of a scalar multiple
Invertibility criterion
Cofactor matrix
Adjugate matrix
Inverse by the adjugate formula
Cramer's rule
Orthogonal matrices
Skew-symmetric matrices
Integer matrices with integer inverses

The essentials

Four rules, one number per matrix

• $\det(AB) = \det A \, \det B$ , in either order, even though $AB \neq BA$ in general.
• $\det(A^{T}) = \det A$ and $\det(A^{m}) = (\det A)^{m}$ .
• $A$ is invertible if and only if $\det A \neq 0$ , and then $\det(A^{-1}) = \dfrac{1}{\det A}$ : the inverse DIVIDES, it does not change the sign.
• $\det(kA) = k^{n}\det A$ for an $n \times n$ matrix, because $kA = (kI)A$ and $\det(kI) = k^{n}$ . One ROW times $k$ gives only $k\det A$ .
• So a composite determinant is one small number per factor: with $3 \times 3$ matrices, $\det A = 2$ and $\det B = -3$ , $\det(2A^{-1}B^{T}) = 8 \times \dfrac{1}{2} \times (-3) = -12$ .

A

triples every area and

B

doubles it, so

BA

multiplies areas by

3 \times 2 = 6

whatever the shapes: that is

\det(BA) = \det B \, \det A

Write $n$ on the line before any scalar rule. The size of the matrix is a hidden input of $\det(kA)$ , $\det(-A)$ and $\det(\text{adj}\,A)$ , and it is the input students forget.

A 2 by 2 determinant is a signed area

• $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ is the area of the parallelogram built on the columns $(a, c)$ and $(b, d)$ , with a sign.
• The sign is $+$ when the turn from the first column to the second is counterclockwise, $-$ when it is clockwise: swapping the two columns flips it.
• Example: $\det\begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} = 5$ , and with the columns swapped $\det\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = -5$ .
• The determinant is $0$ exactly when the columns are parallel: the parallelogram is flat. In $\mathbb{R}^3$ , $|\det A|$ is the volume of the parallelepiped on the columns.

The same two columns give area

5

; the order of the columns gives the sign. Parallel columns, as

(2, 1)

and

(4, 2)

, give no area at all: determinant

0

The adjugate and Cramer: det A in every denominator

• Cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ , with the checkerboard of signs $+ - +$ , $- + -$ , $+ - +$ .
• $\text{adj}(A)$ is the TRANSPOSE of the matrix of cofactors, and $A\,\text{adj}(A) = \text{adj}(A)\,A = \det(A)\,I$ for every square $A$ .
• $A^{-1} = \dfrac{1}{\det A}\,\text{adj}(A)$ , so one entry is $(A^{-1})_{ij} = \dfrac{C_{ji}}{\det A}$ , indices swapped.
• Cramer: if $\det A \neq 0$ , then $x_j = \dfrac{\det A_j(\mathbf{b})}{\det A}$ , where $A_j(\mathbf{b})$ is $A$ with column $j$ replaced by $\mathbf{b}$ .
• $\det(\text{adj}\,A) = (\det A)^{n-1}$ , and an integer matrix has an integer inverse exactly when $\det A = \pm 1$ .

Cramer as a ratio of areas:

b = 2a_1 + a_2

, and replacing

a_1

b

doubles the parallelogram on the columns, from

5

10

. So

x_1 = \frac{\det(b, a_2)}{\det(a_1, a_2)} = \frac{10}{5} = 2

Cramer's rule and one entry of the inverse are the same idea: a single $3 \times 3$ or $2 \times 2$ determinant divided by $\det A$ . When a question asks for ONE unknown or ONE entry, never compute the whole inverse.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

What each new matrix does to the determinant

Read the size first, then the rule. The red line is the one students invent: there is no formula for the determinant of a sum.

New matrix	Size	Determinant
$AB$	any	$\det A \, \det B$
Example: $\det A = 2$ , $\det B = -3$ : $\det(AB) = \det(BA) = -6$ .
$A^{-1}$	any	$\frac{1}{\det A}$
Example: $\det A = -4$ : $\det(A^{-1}) = -\frac{1}{4}$ , same sign, not $4$ .
$kA$	$n \times n$	$k^{n}\det A$
Example: $3 \times 3$ with $\det A = 5$ : $\det(2A) = 40$ , and $80$ if $A$ is $4 \times 4$ .
one row times $k$	any	$k\det A$
Example: $\det D = -4$ : doubling row $1$ gives $-8$ , doubling the $3 \times 3$ matrix gives $-32$ .
$\text{adj}(A)$	$n \times n$	$(\det A)^{n-1}$
Example: $3 \times 3$ with $\det A = 3$ : $\det(\text{adj}\,A) = 9$ .
$A + B$	any	$\det A + \det B$ no such rule
Example: $\det(I_2 + I_2) = \det(2I_2) = 4$ , while $\det I_2 + \det I_2 = 2$ . What to do: Add the matrices first, then take ONE determinant of the sum.

Every line but the last is $\det(AB) = \det A \, \det B$ in disguise: $A^{-1}$ from $AA^{-1} = I$ , $kA$ from $(kI)A$ , $\text{adj}(A)$ from $A\,\text{adj}(A) = \det(A)\,I$ .

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. Forgetting the power n in the determinant of a scalar multiple

2 marks, and every line that uses the value afterwards

What not to write

“ $A$ is $3 \times 3$ with $\det A = 5$ , so $\det(2A) = 2\det A = 10$ .”

What to write

“ $2A$ multiplies all three rows by $2$ , so $\det(2A) = 2^{3}\det A = 40$ .”

In the plane,

3A

stretches every edge by

3

, and the unit square becomes nine unit squares:

\det(3A) = 3^{2}\det A

, never

3\det A

Why: Only a ROW operation multiplies the determinant by $k$ . The scalar in front of the matrix multiplies each of the $n$ rows, which is the figure: three times every edge of a square is nine squares.

2. Letting the scalar escape from, or into, an inverse

the whole part, the two answers differ by a factor of 64

What not to write

“ $\det\big((2A)^{-1}\big) = 2^{3} \cdot \frac{1}{\det A} = \frac{8}{5}$ .”

What to write

“ $\det(2A) = 8 \times 5 = 40$ , so $\det\big((2A)^{-1}\big) = \frac{1}{40}$ ; it is $\det(2A^{-1})$ that equals $\frac{8}{5}$ .”

Why: In $(2A)^{-1} = \frac{1}{2}A^{-1}$ the scalar is inverted along with the matrix; in $2A^{-1}$ it is not. Read the parentheses before choosing a rule.

3. Keeping one branch when the determinant is known through its square

the whole part, with a conclusion opposite to the truth

What not to write

“ $\det(A_k^{T}A_k) = 16$ , so $\det A_k = 4$ ; this has no real solution, so no $k$ works.”

What to write

“ $\det(A_k)^{2} = 16$ , so $\det A_k = 4$ OR $\det A_k = -4$ ; the first branch has no real solution, the second gives $k = 3$ or $k = -2$ .”

Why: $\det(A^{T}A)$ , $\det(\text{adj}\,A)$ for a $3 \times 3$ and $\det(A^{2})$ all give a SQUARE, and a square has two roots. In the series example $\det A_k = (2 - k)(k + 1)$ , which never reaches $4$ and reaches $-4$ twice.

4. Dividing by a determinant that may be zero

2 marks, and the proof is marked as wrong, not incomplete

What not to write

“ $A^{2} = A$ gives $\det(A)^{2} = \det A$ , so $\det A = 1$ .”

What to write

“ $\det A\,(\det A - 1) = 0$ , so $\det A = 0$ or $\det A = 1$ , and $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$ shows that $0$ happens.”

Why: Dividing by $\det A$ silently assumes $A$ invertible, which is not a hypothesis here. Factor instead of dividing, and every root survives.

5. Swapping back the indices of one entry of the inverse

the whole part on a one-entry question

What not to write

“ $(M^{-1})_{23} = \frac{C_{23}}{\det M} = \frac{1}{10}$ .”

What to write

“ $(M^{-1})_{23} = \frac{C_{32}}{\det M} = \frac{-5}{10} = -\frac{1}{2}$ : the adjugate is the TRANSPOSE of the cofactor matrix.”

Why: Row $i$ of the adjugate holds the cofactors of COLUMN $i$ of $M$ . The two values coincide only for a symmetric matrix, which is exactly why a symmetric example in class hides the trap.

6. Reading a zero determinant as no solution

2 marks, and often the rest of the question

What not to write

“ $\det A = 0$ , so by Cramer's rule the system has no solution.”

What to write

“ $\det A = 0$ , so Cramer's rule does not apply; row reduction of the augmented matrix decides between no solution and infinitely many.”

Same coefficients, same determinant

0

: on the left two parallel lines and no solution, on the right one line written twice and infinitely many.

Why: Cramer's rule has a hypothesis, $\det A \neq 0$ . When it fails the rule is silent, and both remaining cases really happen with the same coefficient matrix.

7. Proving a statement about the sign without mentioning the size

the whole proof, since the statement is false for even sizes

What not to write

“ $S^{T} = -S$ gives $\det S = -\det S$ , so every skew-symmetric matrix is singular.”

What to write

“ $\det(-S) = (-1)^{n}\det S$ ; for $n$ odd this gives $\det S = -\det S$ , hence $\det S = 0$ .”

The skew-symmetric

\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

turns the unit square a quarter turn clockwise: same area, determinant

1

, not

0

. Size

2

is even.

Why: $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ is skew-symmetric with determinant $1$ . The step $\det(-S) = -\det S$ is the power $k^{n}$ with $k = -1$ , and it needs $n$ odd.

8. Adding determinants as if they were matrices

2 marks on a true or false question

What not to write

“ $\det A = \det B = 0$ , so $\det(A + B) = 0$ and $A + B$ is singular.”

What to write

“There is no rule for $\det(A + B)$ : here $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = I$ , which is invertible.”

Why: The determinant is multiplicative, not additive. Two flattened shapes can add up to a full square, and $\det(I_2 + I_2) = 4 \neq 1 + 1$ .

Which method to choose

Which tool the question is really asking for

Read what is GIVEN and what is asked, before writing any determinant

If determinants of $A$ and $B$ given, a product of powers, inverses, transposes and scalars asked → one factor per line with the four rules, the scalar raised to the power n

Example: $\det(2A^{-1}B^{T}A^{3}) = 8 \times \frac{1}{2} \times (-3) \times 8 = -96$
If a matrix with a parameter and the words invertible or singular → a factored determinant, then exclude its roots

Example: $\det A_k = (2 - k)(k + 1)$ : invertible for $k \neq 2, -1$
If the whole inverse of a $3 \times 3$ → nine cofactors, TRANSPOSE, divide by the determinant, check one row of the product

Example: $\det A = 2$ : $A^{-1} = \frac{1}{2}\,\text{adj}(A)$
If ONE entry of the inverse → one cofactor with swapped indices, $(A^{-1})_{ij} = C_{ji}/\det A$

Example: $(M^{-1})_{23} = C_{32}/10 = -\frac{1}{2}$
If ONE unknown of a square system with $\det A \neq 0$ → Cramer's rule: replace that column by b, two determinants

Example: $y = \frac{-20}{10} = -2$
If $\det A = 0$ and the question asks how many solutions → row reduction of the augmented matrix; Cramer is silent

Example: $x + y = 2$ , $2x + 2y = 5$ : a row $0 = 1$ , no solution
If prove something about the determinant from a matrix equation → take determinants of both sides, keep every root, mention n

Example: $A^{3} = 2I$ in size $3$ : $\det(A)^{3} = 8$ , so $\det A = 2$

All the unknowns of a large system are cheaper by row reduction than by Cramer's rule: three unknowns cost four $3 \times 3$ determinants. Cramer wins when ONE unknown is asked, or when a parameter sits in the right-hand side.

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.

Evaluating a composite determinant

When to use it: A product of known matrices with inverses, transposes, powers and a scalar, and only $\det A$ , $\det B$ given

1 Write the size $n$ on the first line, because the scalar rule needs it.
2 Split the product with $\det(XY) = \det X \, \det Y$ , one factor per line.
3 Evaluate each factor with its rule: $k^{n}$ for the scalar, $\frac{1}{\det A}$ for an inverse, nothing for a transpose, $(\det A)^{m}$ for a power.
4 Multiply the numbers and simplify, leaving the order of the matrices alone: only their determinants commute.

Concluding sentence

“Since $A$ and $B$ are $3 \times 3$ : $\det(2A^{-1}B^{T}A^{3}) = 2^{3}\det(A^{-1})\det(B^{T})\det(A)^{3} = 8 \times \frac{1}{2} \times (-3) \times 8 = -96$ .”

The trap: Simplifying the MATRIX $A^{-1}B^{T}A^{3}$ into $B^{T}A^{2}$ before taking the determinant: the answer happens to be the same, the line is still marked wrong.

Marking: Usually 1 mark for the scalar power, 1 for the inverse and transpose factors, 1 for the product.

A proof by taking determinants

When to use it: A matrix identity is given ( $A^{2} = A$ , $Q^{T}Q = I$ , $S^{T} = -S$ , $AB = -BA$ ) and a statement about $\det$ is asked

1 Take the determinant of both sides of the given identity.
2 Rewrite each side with the product, transpose and scalar rules, writing $(-1)^{n}$ or $k^{n}$ explicitly.
3 Solve the resulting equation between numbers by FACTORING, never by dividing by the determinant.
4 State every root, then say where the size or the parity of n was used.

Concluding sentence

“Taking determinants, $\det(Q^{T}Q) = \det(Q)^{2} = \det I = 1$ , hence $\det Q = 1$ or $\det Q = -1$ .”

The trap: A proof that never writes $n$ , on a statement that uses $(-1)^{n}$ : it would prove a false statement for even sizes.

Check before you hand in

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

Check a factored determinant at one easy value

After factoring $\det A_k$ , compute the original determinant at $k = 0$ or $k = 1$ directly and compare.

$\det A_0 = 2$ by expansion, and $(2 - 0)(0 + 1) = 2$ : consistent. A $-2$ would reveal a sign error.
One row of A times adj(A)

Before dividing by $\det A$ , multiply one row of $A$ by the three columns of $\text{adj}(A)$ : you must read $\det A$ once and $0$ twice.

$(1, 1, 0)$ against the columns of $\text{adj}(A)$ gives $(2, 0, 0)$ . Getting $(3, \ldots)$ means the transpose was forgotten.
Substitute the Cramer value

Once one unknown is found, an equation with few terms usually gives another, then all equations can be checked.

$z = 3$ and $x + z = 4$ give $x = 1$ ; then $2 - 2 - 3 = -3$ checks the first equation.
Sign and size of a determinant of an inverse

$\det(A^{-1})$ has the SAME sign as $\det A$ , and a large $\det A$ gives a small $\det(A^{-1})$ .

$\det A = -3$ with an answer $\det(A^{-1}) = 3$ is wrong twice, in sign and in size: it is $-\frac{1}{3}$ .

The typical problem, taken apart

A final-exam question, from the parameter to one unknown

For every real $t$ , $N_t = \begin{pmatrix} 1 & 2 & t \\ 0 & t & 1 \\ 1 & -1 & 2 \end{pmatrix}$ .

a) For which $t$ is $N_t$ invertible? b) For $t = 0$ , find the entry in row $1$ , column $2$ of $N_0^{-1}$ . c) Solve $N_0\mathbf{x} = (3, 2, 4)$ for $x_1$ alone. d) Compute $\det(2N_0^{-1})$ .

Step 1

$R_3 \to R_3 - R_1$ leaves the determinant unchanged and gives row $3$ equal to $(0, -3, 2 - t)$ . Expand along column $1$ , where only the top entry $1$ remains: $\det N_t = \det\begin{pmatrix} t & 1 \\ -3 & 2 - t \end{pmatrix} = t(2 - t) + 3 = (3 - t)(t + 1)$ .

Why

Naming the row operation and the column of the expansion earns the method mark; producing two zeros in one column turns a three-term expansion into one term.

Step 2

$N_t$ is invertible if and only if $(3 - t)(t + 1) \neq 0$ , that is for every real $t$ except $t = 3$ and $t = -1$ . For $t = 0$ , $\det N_0 = 3 \neq 0$ .

Why

The conclusion is stated as a condition on t, and the value used later is checked against it before any inverse is written.

Step 3

$(N_0^{-1})_{12} = \dfrac{C_{21}}{\det N_0}$ . Delete row $2$ and column $1$ : $\det\begin{pmatrix} 2 & 0 \\ -1 & 2 \end{pmatrix} = 4$ , sign $(-1)^{2+1} = -1$ , so $C_{21} = -4$ and the entry is $-\dfrac{4}{3}$ .

Why

The indices are swapped on purpose. Using $C_{12} = 1$ would give $\frac{1}{3}$ , and the marker checks precisely this.

Step 4

Replace column $1$ by $(3, 2, 4)$ : $\det\begin{pmatrix} 3 & 2 & 0 \\ 2 & 0 & 1 \\ 4 & -1 & 2 \end{pmatrix} = 3(0 + 1) - 2(4 - 4) + 0 = 3$ , so $x_1 = \dfrac{3}{3} = 1$ .

Why

One unknown costs one extra determinant; the full inverse would have cost nine cofactors for the same answer.

Step 5

$\det(2N_0^{-1}) = 2^{3} \times \dfrac{1}{3} = \dfrac{8}{3}$ .

Why

The scalar multiplies a $3 \times 3$ matrix, so it enters as $2^{3}$ ; writing $\frac{2}{3}$ is the trap of the chapter.

Step 6

Check: row $2$ of the system reads $x_3 = 2$ , row $1$ gives $1 + 2x_2 = 3$ , so $x_2 = 1$ , and row $3$ gives $1 - 1 + 4 = 4$ .

Why

Substituting back takes three lines and validates the determinant, the Cramer numerator and the arithmetic at once.

The conclusion, written out

“ $N_t$ is invertible for $t \neq 3$ and $t \neq -1$ ; for $t = 0$ , $(N_0^{-1})_{12} = -\frac{4}{3}$ , $x_1 = 1$ and $\det(2N_0^{-1}) = \frac{8}{3}$ .”

The classic mistake on this problem: Taking $C_{12}$ instead of $C_{21}$ in c) gives $\frac{1}{3}$ , and writing $\det(2N_0^{-1}) = \frac{2}{3}$ in d): two lost parts on a question that was otherwise right.

Learn by heart

• $\det(AB) = \det A \, \det B$ , $\det(A^{T}) = \det A$ , $\det(A^{-1}) = \dfrac{1}{\det A}$ .
• $\det(kA) = k^{n}\det A$ and $\det(-A) = (-1)^{n}\det A$ : write $n$ FIRST.
• $A$ invertible $\iff \det A \neq 0$ .
• $\text{adj}(A) = [C_{ij}]^{T}$ , $A\,\text{adj}(A) = \det(A)\,I$ , $A^{-1} = \dfrac{\text{adj}(A)}{\det A}$ .
• One entry: $(A^{-1})_{ij} = \dfrac{C_{ji}}{\det A}$ , indices SWAPPED.
• Cramer: $x_j = \dfrac{\det A_j(\mathbf{b})}{\det A}$ , only when $\det A \neq 0$ ; otherwise row reduce.
• $\det(\text{adj}\,A) = (\det A)^{n-1}$ ; integer inverse $\iff \det A = \pm 1$ .
• No rule for $\det(A + B)$ . Never.

Frequently asked questions

Why is the determinant of 2A not twice the determinant of A?

Because multiplying the matrix by 2 multiplies every one of its n rows by 2, and each row contributes one factor of 2. For a 3 by 3 matrix the determinant is multiplied by 8, for a 4 by 4 by 16. Only multiplying a single row by 2 doubles the determinant. Geometrically, doubling every edge of a cube multiplies its volume by 8.

How do I find one entry of an inverse matrix without computing the whole inverse?

Divide one cofactor by the determinant, with the indices swapped: the entry in row i and column j of the inverse is the cofactor of position j, i of the original matrix, divided by its determinant. The swap comes from the transpose in the adjugate. It costs one 2 by 2 determinant instead of nine.

When should I use Cramer's rule instead of row reduction?

When the determinant of the coefficient matrix is not zero and only one unknown is asked, or when a parameter appears in the right-hand side. One unknown then costs two determinants. For all the unknowns of a system, row reduction is faster, and when the determinant is zero Cramer's rule does not apply at all.

Does a zero determinant mean the system has no solution?

No. A zero determinant means the system does not have exactly one solution: it has either none or infinitely many, and row reduction of the augmented matrix decides which. Two parallel lines and one line written twice both have a zero determinant, and only the first case has no solution.

Practise it

Corrected exercises: Determinants, inverses and Cramer's rule, 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

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Revision sheet: determinants, inverses and Cramer's rule (MATH 133)

The essentials

Four rules, one number per matrix

A 2 by 2 determinant is a signed area

The adjugate and Cramer: det A in every denominator

The rules in table form

What each new matrix does to the determinant

The mistakes that cost marks

1. Forgetting the power n in the determinant of a scalar multiple

2. Letting the scalar escape from, or into, an inverse

3. Keeping one branch when the determinant is known through its square

4. Dividing by a determinant that may be zero

5. Swapping back the indices of one entry of the inverse

6. Reading a zero determinant as no solution

7. Proving a statement about the sign without mentioning the size

8. Adding determinants as if they were matrices

Which method to choose

Which tool the question is really asking for

How the answer is expected to be written

Evaluating a composite determinant

A proof by taking determinants

Check before you hand in

The typical problem, taken apart

A final-exam question, from the parameter to one unknown

Learn by heart

Frequently asked questions

Why is the determinant of 2A not twice the determinant of A?

How do I find one entry of an inverse matrix without computing the whole inverse?

When should I use Cramer's rule instead of row reduction?

Does a zero determinant mean the system has no solution?

Corrected exercises: Determinants, inverses and Cramer's rule, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?