MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: matrix inverses and elementary matrices (MATH 133)

This sheet is not a summary of sections 2.4 and 2.5 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on matrix inverses and elementary matrices in MATH 133 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam: the $2 \times 2$ inverse by its formula, every larger inverse by row reducing $[A \mid I]$ with each operation named, and every answer checked by multiplying back. Determinants of $3 \times 3$ matrices, the adjugate and Cramer's rule belong to the next chapters and are not needed here.

The thread of the chapter

There is no matrix division: an inverse undoes a matrix from ONE side, and the side is the whole answer. Every row operation multiplies on the left, $(AB)^{-1}$ reverses the order, $AX = B$ and $XA = B$ have different solutions, and a product of elementary matrices is read from the side it acts on.

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

Inverse of a matrix
Invertible matrix
Uniqueness of the inverse
Inverse of a 2x2 matrix
Matrix inversion algorithm
Gauss-Jordan elimination
Inverse of a product
Inverse of a transpose
Matrix equations
Elementary matrices
Product of elementary matrices
Nilpotent matrix

The essentials

Definition, uniqueness, and the side

• $A$ (square) is invertible when some $B$ gives $AB = I$ AND $BA = I$ . That $B$ is unique, because $B = B(AC) = (BA)C = C$ , and it is written $A^{-1}$ .
• For square matrices one product is enough: $AB = I$ already forces $BA = I$ . On a first question, write both anyway.
• $2 \times 2$ : $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ when $ad - bc \neq 0$ ; no inverse when $ad - bc = 0$ .
• There is no $\frac{B}{A}$ . $AX = B$ gives $X = A^{-1}B$ , $XA = B$ gives $X = BA^{-1}$ , $A\mathbf{x} = \mathbf{b}$ gives $\mathbf{x} = A^{-1}\mathbf{b}$ .

Same letters, two equations, two answers:

A^{-1}

enters on the side where

A

stood, so

A^{-1}B

and

BA^{-1}

are different matrices in general.

Every question of the chapter reduces to one habit: before writing $A^{-1}$ anywhere, say on which side of the unknown $A$ stands, and multiply BOTH sides of the equation on that side.

The algorithm, and what the right block records

• Row reduce $[A \mid I]$ . If the left block becomes $I$ , the right block is $A^{-1}$ . If a ROW of zeros appears in the left block, stop: $A$ has no inverse.
• Each row operation is a multiplication on the LEFT by an elementary matrix $E$ , the matrix obtained by doing that operation to $I$ .
• If $E_k \cdots E_2E_1A = I$ , then $A^{-1} = E_k \cdots E_2E_1$ , which is exactly what the right block holds, and $A = E_1^{-1}E_2^{-1} \cdots E_k^{-1}$ .
• Inverses of elementary matrices: a swap is its own inverse, scaling by $c \neq 0$ is undone by scaling by $\frac{1}{c}$ , adding $cR_j$ is undone by subtracting $cR_j$ .
• Rules: $(AB)^{-1} = B^{-1}A^{-1}$ , $(A^T)^{-1} = (A^{-1})^T$ , $(kA)^{-1} = \frac{1}{k}A^{-1}$ , $(A^n)^{-1} = (A^{-1})^n$ . None for $(A + B)^{-1}$ .

Column operations are multiplications on the RIGHT: one of them in the middle of the algorithm and the right block is no longer $A^{-1}$ .

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.

Inverting a combination of matrices

Read the expression, then the tempting answer students write, then the correct inverse. Throughout, $A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ , so $A^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$ and $B^{-1} = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$ . The red line is the one rule that does not exist.

Expression	Tempting answer	Correct inverse
$(AB)^{-1}$	$A^{-1}B^{-1}$	$B^{-1}A^{-1}$
Example: $AB = \begin{pmatrix} 4 & 1 \\ 7 & 2 \end{pmatrix}$ and $(AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 2 & -1 \\ -7 & 4 \end{pmatrix}$ , while $A^{-1}B^{-1} = \begin{pmatrix} 5 & -1 \\ -4 & 1 \end{pmatrix}$ .
$(B^T)^{-1}$	$B^{-1}$	$(B^{-1})^T$
Example: $B^T = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ has inverse $\begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ , the transpose of $B^{-1}$ .
$(2B)^{-1}$	$2B^{-1}$	$\frac{1}{2}B^{-1}$
Example: $2B \cdot \frac{1}{2}B^{-1} = I$ , while $2B \cdot 2B^{-1} = 4I$ .
$(B^3)^{-1}$	$\frac{1}{B^3}$	$(B^{-1})^3$
Example: $B^3 = \begin{pmatrix} 1 & 0 \\ 9 & 1 \end{pmatrix}$ and $(B^{-1})^3 = \begin{pmatrix} 1 & 0 \\ -9 & 1 \end{pmatrix}$ : their product is $I$ .
$(A + B)^{-1}$	$A^{-1} + B^{-1}$	no formula rule does not exist
Example: $I + I = 2I$ has inverse $\frac{1}{2}I$ , but $I^{-1} + I^{-1} = 2I$ . Worse, $I + (-I) = 0$ has no inverse at all. What to do: Add first, check that the sum is invertible, then invert the sum like any other matrix.

Every correct line of this table is proved the same way: name the candidate, multiply it by the matrix on both sides, get $I$ , and conclude by uniqueness of the inverse.

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. Solving XA = B as if A stood on the left

the whole question

What not to write

“ $XA = B$ , so $X = A^{-1}B$ .”

What to write

“ $A$ is on the right of $X$ , so I multiply both sides on the right by $A^{-1}$ : $X = BA^{-1}$ .”

Why: With $A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$ , $BA^{-1} = \begin{pmatrix} 2 & -1 \\ 5 & -2 \end{pmatrix}$ solves the equation and $A^{-1}B = \begin{pmatrix} -1 & -1 \\ 2 & 1 \end{pmatrix}$ does not. Check sizes first: if $B$ has three columns and $A$ is $2 \times 2$ , $XA = B$ has no solution of any size.

2. Keeping the original order in the inverse of a product

2 marks, and every result built on it

What not to write

“ $(AB)^{-1} = A^{-1}B^{-1}$ .”

What to write

“ $(AB)^{-1} = B^{-1}A^{-1}$ : $B$ acted first, so it is undone last.”

Why: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = I$ , and the other order leaves $A$ and $A^{-1}$ separated by $B$ . The same reversal governs $A = E_1^{-1}E_2^{-1}E_3^{-1}$ and $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ .

3. Inverting a matrix entry by entry

the whole question

What not to write

“ $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{4} \end{pmatrix}$ .”

What to write

“ $ad - bc = 4 - 6 = -2$ , so the inverse is $-\frac{1}{2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$ .”

Why: The $(1,1)$ entry of the product of the matrix with the false answer is $1 + \frac{2}{3} = \frac{5}{3}$ , not $1$ . An inverse undoes the whole map, so each of its entries depends on all four entries of $A$ .

4. Swapping the wrong entries in the 2 by 2 formula

2 marks

What not to write

“ $\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix}$ .”

What to write

“Swap the DIAGONAL, change the SIGN of the other two, divide by $ad - bc = 1$ : $\begin{pmatrix} 1 & -3 \\ -1 & 4 \end{pmatrix}$ .”

Why: The false answer gives $\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 24 \\ 2 & 7 \end{pmatrix}$ , visibly not $I$ . Four entries, two gestures: the diagonal entries trade places, the off-diagonal ones stay in place and change sign.

5. Row reducing the left block and forgetting the right one

the whole question, even with a correct left block

What not to write

“ $R_2 \to R_2 - 2R_1$ on $A$ first, and the identity copied on the right at the end.”

What to write

“Every operation is applied to the WHOLE row of $[A \mid I]$ , left and right blocks together.”

Why: The right block is the record of the operations: $E_k \cdots E_1[A \mid I] = [I \mid E_k \cdots E_1]$ . An operation skipped on the right is missing from the product, so the matrix read at the end is not $A^{-1}$ , and multiplying back shows it.

6. Declaring a matrix singular because of a zero in the pivot place

the whole question

What not to write

“The $(1,1)$ entry of $\begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix}$ is $0$ , so the matrix is not invertible.”

What to write

“ $R_1 \leftrightarrow R_2$ brings a $2$ into the pivot place and the reduction goes on: the inverse is $\begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \\ 1 & 0 \end{pmatrix}$ .”

Left: a

0

in the pivot place with a

1

below it, so swap and continue. Right: a whole row of zeros in the left block, the only situation that proves there is no inverse.

Why: A zero in the pivot place calls for a look BELOW it and a swap. Only a whole row of zeros in the left block ends the algorithm, which happens when the pivot column has nothing but zeros at and below the pivot place.

7. Dividing by a pivot that depends on a letter

half the question, the case analysis

What not to write

“For $\begin{pmatrix} 1 & 2 \\ 3 & k \end{pmatrix}$ , $R_2 - 3R_1$ gives the pivot $k - 6$ , then $R_2 \to \frac{1}{k-6}R_2$ : invertible for every $k$ .”

What to write

“If $k = 6$ , the second row is zero and there is no inverse. If $k \neq 6$ , the pivot is non-zero and I may divide by it: invertible exactly when $k \neq 6$ .”

Why: Dividing by $k - 6$ silently assumes $k \neq 6$ . The excluded value is the whole answer to for which $k$ , so it must be stated BEFORE the division, as its own case.

8. Writing A as the product of the elementary matrices in the order found

the whole part

What not to write

“ $E_3E_2E_1A = I$ , so $A = E_1E_2E_3$ .”

What to write

“ $E_3E_2E_1A = I$ , so $A^{-1} = E_3E_2E_1$ and $A = E_1^{-1}E_2^{-1}E_3^{-1}$ .”

Why: Going back from $I$ to $A$ means UNDOING the operations, so the factors are the inverses, and the last operation performed is undone first. With the three operations taking $\begin{pmatrix} 0 & 2 \\ 1 & 3 \end{pmatrix}$ to $I$ , the false product is $\begin{pmatrix} 0 & \frac{1}{2} \\ 1 & -3 \end{pmatrix}$ , another matrix.

9. Factoring X out without the identity

1 mark, and a meaningless line

What not to write

“ $3X - MX = D$ , so $(3 - M)X = D$ .”

What to write

“ $3X - MX = (3I - M)X = D$ , so $X = (3I - M)^{-1}D$ when $3I - M$ is invertible.”

Why: A number minus a matrix is not defined: $3X$ is $(3I)X$ , and the $I$ is what makes the factorisation legal. Factor $X$ out on the side where it stands in every term, here the right.

Which method to choose

Which method for the inverse, by the form of the matrix

Look at the size and the shape of the matrix, and at what the statement gives you

For a

2 \times 2

ad - bc

decides: non-zero, the two lines of

A\mathbf{x} = \mathbf{b}

cross once for every

\mathbf{b}

; zero, the rows are proportional and the lines are parallel.

If a numerical $2 \times 2$ → the formula with $ad - bc$

Example: $\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -3 \\ -1 & 4 \end{pmatrix}$
If a numerical $3 \times 3$ or larger → row reduce $[A \mid I]$ , naming every operation

Example: the worked example below, four operations
If a letter inside the matrix → reduce with the letter, isolate the pivot that contains it, and split the cases BEFORE dividing

Example: $\begin{pmatrix} 1 & 2 \\ 3 & k \end{pmatrix}$ : pivot $k - 6$ , invertible iff $k \neq 6$
If a diagonal matrix → invert each diagonal entry

Example: $\text{diag}(2, -1, 4)^{-1} = \text{diag}(\frac{1}{2}, -1, \frac{1}{4})$
If $I - N$ with $N$ strictly triangular → $(I - N)^{-1} = I + N + N^2 + \dots$ , which stops because a power of $N$ is zero

Example: $N^2 = 0$ gives $(I - N)^{-1} = I + N$
If a polynomial identity satisfied by A → factor $A$ out on each side to get $A \cdot (\dots) = I$

Example: $A^2 - 2A - 3I = 0$ gives $A^{-1} = \frac{1}{3}(A - 2I)$
If a product, a transpose or a power of matrices whose inverses are known → use the rules of the table, never a new reduction

Example: $(AB)^{-1} = B^{-1}A^{-1}$

Not in this chapter: a $3 \times 3$ determinant to decide invertibility, the adjugate, Cramer's rule. They come next, and a solution built on them before they are taught earns the method marks of none of the above.

Which side: isolating X in a matrix equation

Look where each known matrix stands relative to X

If $AX = B$ → multiply on the LEFT by $A^{-1}$

Example: $X = A^{-1}B$
If $XA = B$ → multiply on the RIGHT by $A^{-1}$

Example: $X = BA^{-1}$
If $PXQ = R$ → peel from the outside, left factor from the left, right factor from the right

Example: $X = P^{-1}RQ^{-1}$
If an inverse around the unknown, $(AX)^{-1} = N$ → invert both sides first, then peel

Example: $X = A^{-1}N^{-1} = (NA)^{-1}$
If $X$ in several terms, $kX - MX = D$ → factor $X$ on its side with the identity

Example: $X = (kI - M)^{-1}D$

If no branch fits, compare the sizes: the product on each side of the equation must have the size of the other side, and that alone usually tells you where X must go.

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.

Inverting a matrix by row reduction

When to use it: Any statement that says find $A^{-1}$ for a $3 \times 3$ , or decide whether $A$ is invertible and if so find its inverse

1 Write the block $[A \mid I]$ in full, with the vertical bar.
2 Reduce, naming every operation in the margin: $R_2 \to R_2 - 2R_1$ , $R_1 \leftrightarrow R_3$ , $R_3 \to \frac{1}{2}R_3$ . Apply each one to the whole row.
3 If a whole row of the left block becomes zero, stop and conclude that A is not invertible, quoting that row.
4 When the left block is $I$ , copy the right block as $A^{-1}$ .
5 Multiply back: compute $AA^{-1}$ , or at least one row of it, and write that you obtained $I$ .

Concluding sentence

“The left block reduced to $I$ , so $A$ is invertible and $A^{-1}$ is the right block. Check: $AA^{-1} = I$ .”

The trap: Stopping at the matrix with no sentence and no check. The marker cannot tell a copied answer from a computed one, and the check is where a sign error becomes visible.

Marking: Typically 1 mark for the set-up, 3 for the operations named and applied correctly, 1 for reading the answer and 1 for the check. The named operations are worth more than the final matrix.

Proving invertibility from a polynomial identity

When to use it: A statement that gives an identity such as $A^2 - 2A - 3I = 0$ and asks to show that $A$ is invertible

1 Isolate the identity term: $A^2 - 2A = 3I$ .
2 Factor $A$ out on the LEFT: $A(A - 2I) = 3I$ , then on the RIGHT: $(A - 2I)A = 3I$ .
3 Divide the NUMBER, never a matrix: $A \cdot \frac{1}{3}(A - 2I) = I$ and $\frac{1}{3}(A - 2I) \cdot A = I$ .
4 Conclude by the definition and the uniqueness of the inverse.

Concluding sentence

“Since $A \cdot \frac{1}{3}(A - 2I) = \frac{1}{3}(A - 2I) \cdot A = I$ , $A$ is invertible and $A^{-1} = \frac{1}{3}(A - 2I)$ .”

The trap: Writing $A = \frac{3I}{A - 2I}$ , a fraction of matrices that does not exist, or factoring the identity as $(A - 3I)(A + I) = 0$ and concluding $A = 3I$ or $A = -I$ : a product of two non-zero matrices can be zero.

Marking: Usually 2 marks for the factorisation on both sides and 1 for the conclusion that names the inverse.

Check before you hand in

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

Multiply back one column

The product of $A$ with column $j$ of your answer must be column $j$ of $I$ . One column costs three dot products for a $3 \times 3$ and catches most slips.

With the worked example below, $A$ times the first column $(0, 1, -1)$ gives $(0 + 2 - 1, 0 + 1 - 1, 0 + 2 - 2) = (1, 0, 0)$ . A result such as $(1, 0, 2)$ says a row operation went wrong in the third row.
The shape survives inversion

The inverse of an upper triangular matrix is upper triangular with the reciprocal diagonal; the inverse of a symmetric matrix is symmetric; the inverse of a diagonal matrix is diagonal.

$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ . An answer with a non-zero entry below the diagonal would be wrong on sight.
For a $2 \times 2$ , the reciprocal number

The number $ad - bc$ of $A^{-1}$ is the reciprocal of that of $A$ .

$\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ has $-2$ , its inverse $\begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$ has $1 - \frac{3}{2} = -\frac{1}{2}$ .
Sizes before sides

Before solving a matrix equation, check that the product you are about to write exists and has the size of the other side.

If $A$ is $2 \times 2$ and $B$ is $2 \times 3$ , then $X = A^{-1}B$ exists for $AX = B$ , and $XA = B$ has no solution at all, since $XA$ always has $2$ columns.

The typical problem, taken apart

An inverse by row reduction, then an equation solved on the correct side

Let $A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 2 \end{pmatrix}$ .

Find $A^{-1}$ by reducing $[A \mid I]$ , then find the row vector $\mathbf{x}$ such that $\mathbf{x}A = (2, 5, 4)$ , and say what $\mathbf{x}$ means for the rows of $A$ .

Step 1

$[A \mid I] = \left(\begin{array}{ccc|ccc} 1 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 2 & 2 & 0 & 0 & 1 \end{array}\right)$ . $R_3 \to R_3 - R_1$ gives the third row $(0, 0, 1 \mid -1, 0, 1)$ .

Why

One operation already produces a staircase with pivots 1, 1, 1: the matrix is invertible, and the rest is clearing upwards. Spotting the cheapest first move is worth more than following a fixed recipe.

Step 2

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

Why

Clearing the third column with the third row first means the later operations never touch it again. Every operation is applied to the whole row, right block included, because the right block is the record.

Step 3

$R_1 \to R_1 - 2R_2$ gives $(1, 0, 0 \mid 0, -2, 1)$ . The block is $\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -2 & 1 \\ 0 & 1 & 0 & 1 & 1 & -1 \\ 0 & 0 & 1 & -1 & 0 & 1 \end{array}\right)$ , so $A^{-1} = \begin{pmatrix} 0 & -2 & 1 \\ 1 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}$ .

Why

Four named operations: that list is where most of the marks are, whatever happens to the final matrix.

Step 4

Check: the rows of $A$ against the columns of $A^{-1}$ give $1, 0, 0$ , then $0, 1, 0$ , then $0, 0, 1$ . So $AA^{-1} = I$ .

Why

Nine dot products, one minute. A sign error in the reduction shows up here as a wrong entry, never silently.

Step 5

$\mathbf{x}A = (2, 5, 4)$ has $A$ on the RIGHT of the unknown, so $\mathbf{x} = (2, 5, 4)A^{-1} = (0 + 5 - 4,\ -4 + 5 + 0,\ 2 - 5 + 4) = (1, 1, 1)$ .

Why

A row vector times a matrix is a combination of its ROWS: $\mathbf{x}A = (2, 5, 4)$ asks which combination of the rows of $A$ gives $(2, 5, 4)$ , and $(1, 1, 1)$ says the plain sum. Check: $(1, 2, 1) + (0, 1, 1) + (1, 2, 2) = (2, 5, 4)$ .

The conclusion, written out

“ $A^{-1} = \begin{pmatrix} 0 & -2 & 1 \\ 1 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}$ , and $\mathbf{x} = (2, 5, 4)A^{-1} = (1, 1, 1)$ : the sum of the three rows of $A$ is $(2, 5, 4)$ .”

The classic mistake on this problem: Computing $A^{-1}$ times the column $(2, 5, 4)$ instead: that gives $(-6, 3, 2)$ , which solves $A\mathbf{x} = \mathbf{b}$ , a different equation, and $-6R_1 + 3R_2 + 2R_3 = (-4, -5, 1)$ , nowhere near $(2, 5, 4)$ .

Learn by heart

• $A^{-1}$ exists when $AB = BA = I$ for some $B$ , and it is unique.
• $2 \times 2$ : swap the diagonal, change the sign of the other two, divide by $ad - bc$ ; no inverse when $ad - bc = 0$ .
• $[A \mid I] \to [I \mid A^{-1}]$ ; a zero ROW in the left block means no inverse, a zero ENTRY means swap.
• $AX = B \Rightarrow X = A^{-1}B$ ; $XA = B \Rightarrow X = BA^{-1}$ . Never $\frac{B}{A}$ .
• $(AB)^{-1} = B^{-1}A^{-1}$ , $(A^T)^{-1} = (A^{-1})^T$ , $(kA)^{-1} = \frac{1}{k}A^{-1}$ . No rule for $(A + B)^{-1}$ .
• $E_k \cdots E_1A = I \Rightarrow A^{-1} = E_k \cdots E_1$ and $A = E_1^{-1} \cdots E_k^{-1}$ .
• $N^2 = 0 \Rightarrow (I - N)^{-1} = I + N$ ; $N^3 = 0 \Rightarrow (I - N)^{-1} = I + N + N^2$ .

Frequently asked questions

How do I find the inverse of a 3 by 3 matrix without a calculator?

Write the matrix and the identity side by side and row reduce the whole block until the left half is the identity. The right half is then the inverse. Name every row operation, apply it to the full row, and finish by multiplying the matrix by your answer to check that you get the identity. If a row of zeros appears on the left, the matrix has no inverse.

How do I know a matrix has no inverse in MATH 133?

For a 2 by 2, compute a times d minus b times c: if it is zero, there is no inverse. For anything larger, start the row reduction of the matrix next to the identity. If a whole row of the left half becomes zero, stop: the matrix is not invertible. A single zero in a pivot position is not enough, a row swap usually fixes it.

Is the inverse of AB equal to the inverse of A times the inverse of B?

No, the order reverses: the inverse of AB is the inverse of B times the inverse of A. The matrix B acted first, so it has to be undone last, just as the last step of any procedure is the first one to be undone. The original order gives a different matrix in general, and multiplying it by AB does not give the identity.

What is the difference between AX = B and XA = B?

The side on which A stands. In AX equals B, multiply both sides on the left by the inverse of A, which gives X equals the inverse of A times B. In XA equals B, multiply on the right, which gives X equals B times the inverse of A. Matrices do not commute, so the two answers are different matrices, and only one of them solves each equation.

Why does row reducing next to the identity give the inverse?

Each row operation is the same as multiplying on the left by an elementary matrix. When the left half reaches the identity, the product of all those elementary matrices times A equals the identity, so that product is the inverse of A. The identity on the right started as a blank record and ends up holding exactly that product.

Practise it

Corrected exercises: Matrix inverses and elementary matrices, 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

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Revision sheet: matrix inverses and elementary matrices (MATH 133)

The essentials

Definition, uniqueness, and the side

The algorithm, and what the right block records

The rules in table form

Inverting a combination of matrices

The mistakes that cost marks

1. Solving XA = B as if A stood on the left

2. Keeping the original order in the inverse of a product

3. Inverting a matrix entry by entry

4. Swapping the wrong entries in the 2 by 2 formula

5. Row reducing the left block and forgetting the right one

6. Declaring a matrix singular because of a zero in the pivot place

7. Dividing by a pivot that depends on a letter

8. Writing A as the product of the elementary matrices in the order found

9. Factoring X out without the identity

Which method to choose

Which method for the inverse, by the form of the matrix

Which side: isolating X in a matrix equation

How the answer is expected to be written

Inverting a matrix by row reduction

Proving invertibility from a polynomial identity

Check before you hand in

The typical problem, taken apart

An inverse by row reduction, then an equation solved on the correct side

Learn by heart

Frequently asked questions

How do I find the inverse of a 3 by 3 matrix without a calculator?

How do I know a matrix has no inverse in MATH 133?

Is the inverse of AB equal to the inverse of A times the inverse of B?

What is the difference between AX = B and XA = B?

Why does row reducing next to the identity give the inverse?

Corrected exercises: Matrix inverses and elementary matrices, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?