Multivariable-Calculus · Unit 3 · Video 5 · Interactive Practice

The Matrix That Undoes A: A Inverse, and Why It Multiplies on the Left

IKey Formulas

Formula	Name	What you need
$AA^{-1} = I \qquad A^{-1}A = I$	Definition of the inverse	A square matrix $A$
$A$ of size $n \times n$	The square requirement	As many rows as columns
$AX = B \;\Longrightarrow\; X = A^{-1}B$	Solving a linear system	$A^{-1}$ and the column $B$
$(n \times n)(n \times 1) = (n \times 1)$	The size rule	$AX$ exists, $XA$ does not

Key Insight: $A^{-1}$ multiplies both sides on the left because that is where the $A$ it cancels sits: $A^{-1}(AX) = (A^{-1}A)X = IX = X$ .

IIVisualization 1 — What It Means to Undo A

Applying $A$ and then $A^{-1}$ returns every vector to exactly where it started.

💡 Challenge: find the $X$ that $A$ sends to $\langle 5, 3 \rangle$ — that single question is the system $AX = \langle 5, 3 \rangle$ , solved below.

IIIVisualization 2 — Why the Inverse Goes on the Left

One factor, applied to the left of both sides, turns $AX = B$ into its own solution.

Step 1

IVVisualization 3 — Which Products Even Exist

A square matrix and a column multiply in one order only — the other has no size.

VQuiz Questions

Problem 1 · Which Matrix Is the Inverse

Given: $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ — find the matrix $M$ with $AM = MA = I$ .

✅ Correct! On the diagonal

2 \cdot 1 + 1 \cdot (-1) = 1

, off it

2 \cdot (-1) + 1 \cdot 2 = 0

, so

AM = MA = I

❌ Those are the reciprocals of the entries. Inverting a matrix is not inverting its entries: this

M

gives

AM = \begin{pmatrix} 2 & 3 \\ \tfrac{3}{2} & 2 \end{pmatrix}

❌ Every sign is flipped. This matrix gives

AM = -I

, and

-I

is not the identity.

❌ The diagonal entries have to trade places too. As written,

AM = \begin{pmatrix} 3 & -1 \\ 1 & 0 \end{pmatrix}

❌ Not quite. Test your candidate: multiply it by

A

and see whether every diagonal entry is

1

and every other entry

0

Show solution

For a $2 \times 2$ matrix, swap the diagonal entries, negate the off-diagonal ones, and divide by the determinant:

\det A = 2 \cdot 1 - 1 \cdot 1 = 1, \qquad A^{-1} = \frac{1}{1}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}

Check both products, since either one is enough for a square matrix:

AA^{-1} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 - 1 & -2 + 2 \\ 1 - 1 & -1 + 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

A^{-1}A = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 - 1 & 1 - 1 \\ -2 + 2 & -1 + 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

The exponent $-1$ is a name for the undoing matrix, not an instruction to take reciprocals.

Problem 2 · Solve the System

Given: $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$ , $\;A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$ , $\;B = \langle 5, 9 \rangle$ — find the column $X$ with $AX = B$ .

✅ Correct!

X = A^{-1}B = \langle 2 \cdot 5 + (-1) \cdot 9,\; (-5) \cdot 5 + 3 \cdot 9 \rangle = \langle 1, 2 \rangle

, and

A\langle 1, 2 \rangle = \langle 5, 9 \rangle

❌ That is $AB$ , not $A^{-1}B$ . Multiplying by

A

applies the transformation again instead of undoing it.

❌ Right numbers, wrong order. The first row of

A^{-1}

produces

x_1

2 \cdot 5 + (-1) \cdot 9 = 1

❌ That is $B$ treated as a row and multiplied from the left. With

B

a column,

BA^{-1}

(2 \times 1)(2 \times 2)

and does not exist.

❌ Not quite. Multiply

A^{-1}

into

B

row by row: each row of

A^{-1}

dotted with

B

gives one entry of

X

Show solution

Left-multiply $AX = B$ by $A^{-1}$ ; the left side collapses to $X$ , so

X = A^{-1}B = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 5 \\ 9 \end{pmatrix} = \begin{pmatrix} 10 - 9 \\ -25 + 27 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}

Verify in the original system:

3 \cdot 1 + 1 \cdot 2 = 5 \qquad 5 \cdot 1 + 2 \cdot 2 = 9

Both rows check, so $X = \langle 1, 2 \rangle$ .

Problem 3 · From System to Solution

Given: the system $2x_1 + x_2 = 7$ and $x_1 + x_2 = 4$ , written as $AX = B$ with $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \langle 7, 4 \rangle$ .

What is $A^{-1}$ ?

What is $X$ ?

✅ Correct!

X = A^{-1}B = \langle 7 - 4,\; -7 + 8 \rangle = \langle 3, 1 \rangle

, and indeed

2 \cdot 3 + 1 = 7

and

3 + 1 = 4

❌ Check the inverse. Swap the two diagonal entries, negate the two off-diagonal ones, then divide by

\det A = 2 \cdot 1 - 1 \cdot 1 = 1

❌ Check the multiplication.

x_1

is the first row of

A^{-1}

dotted with

B

1 \cdot 7 + (-1) \cdot 4

❌ The off-diagonal entries have to be negated. Swapping alone gives a matrix

M

with

AM = \begin{pmatrix} 3 & 4 \\ 2 & 3 \end{pmatrix}

❌ That is $A$ itself.

A^2 = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}

, which is not the identity.

❌ The diagonal entries must trade places too. The

2

belongs in the lower-right corner of

A^{-1}

❌ Right numbers, wrong order. The first row of

A^{-1}

builds

x_1

1 \cdot 7 + (-1) \cdot 4 = 3

❌ That is $AB$ , not $A^{-1}B$ . Multiplying by

A

applies the change of variables again instead of reversing it.

❌ That is what a sign-less inverse gives. With the true first row

(1, -1)

x_1 = 7 - 4 = 3

, not

7 + 4

Show solution

Step 1 — invert $A$ . $\det A = 2 \cdot 1 - 1 \cdot 1 = 1$ , so

A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}, \qquad A^{-1}A = \begin{pmatrix} 2 - 1 & 1 - 1 \\ -2 + 2 & -1 + 2 \end{pmatrix} = I

Step 2 — left-multiply the system.

X = A^{-1}B = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 7 \\ 4 \end{pmatrix} = \begin{pmatrix} 7 - 4 \\ -7 + 8 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}

Step 3 — verify both equations.

2 \cdot 3 + 1 = 7 \qquad 3 + 1 = 4

So $x_1 = 3$ and $x_2 = 1$ . Dropping the minus signs in $A^{-1}$ would have given $\langle 11, 1 \rangle$ , and multiplying by $A$ instead of $A^{-1}$ would have given $\langle 18, 11 \rangle$ — neither satisfies the system.

Problem 4 · Multiplying on the Wrong Side

Given: $A$ is an invertible $3 \times 3$ matrix and $X$ , $B$ are $3 \times 1$ columns. Starting from $AX = B$ , a student multiplies both sides on the right by $A^{-1}$ and writes $AXA^{-1} = BA^{-1}$ — what goes wrong first?

✅ Correct!

BA^{-1}

(3 \times 1)(3 \times 3)

and fails the same way. The only legal move is

A^{-1}

on the left of both sides.

❌ There is nothing to collapse.

BA^{-1}

puts a

3 \times 1

column in front of a

3 \times 3

matrix, and that product does not exist.

❌ Close — that is the objection when $X$ is square. Then

AXA^{-1}

exists but does not simplify. Here

X

is a column, so the product never gets that far.

❌ For a square matrix both products are the identity.

AA^{-1} = A^{-1}A = I

; the trouble here is the sizes.

❌ Not quite. Write the sizes under each factor before anything else:

(3 \times 3)(3 \times 1)

works,

(3 \times 1)(3 \times 3)

does not.

Show solution

Check the sizes on the left. $AX$ is $(3 \times 3)(3 \times 1) = 3 \times 1$ , a column. Multiplying it by $A^{-1}$ on the right gives

(3 \times 1)(3 \times 3): \qquad 1 \neq 3

so $(AX)A^{-1}$ is undefined. The right side fails identically: $BA^{-1}$ is also $(3 \times 1)(3 \times 3)$ .

Left-multiplication works because the $A^{-1}$ lands immediately beside the $A$ it cancels:

A^{-1}(AX) = (A^{-1}A)X = IX = X, \qquad X = A^{-1}B

Even if $X$ were square, so that $AXA^{-1}$ existed, the $X$ sitting between $A$ and $A^{-1}$ would keep them apart and nothing would cancel. The rule: multiply by $A^{-1}$ on the left, and on the left of both sides.

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