Math Core

To solve $5x = 15$ you multiply both sides by $5^{-1}$ . For a matrix equation $A\mathbf{x} = \mathbf{b}$ you would like to do the same thing: multiply by a matrix that undoes $A$ . This lesson defines that matrix, gives a formula for it in the $2 \times 2$ case, and shows how row reduction produces it in general.

Invertible matrices

Definition

Inverse of a matrix

An $n \times n$ matrix $A$ is invertible if there is an $n \times n$ matrix $C$ with

CA = I_n \quad \text{and} \quad AC = I_n.

Then $C$ is called the inverse of $A$ and is written $A^{-1}$ . A square matrix that is not invertible is called singular; an invertible matrix is also called nonsingular.

The inverse, when it exists, is unique: if $B$ and $C$ are both inverses of $A$ , then $B = BI = B(AC) = (BA)C = IC = C$ . Only square matrices get inverses in this sense.

The 2 × 2 case

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , multiply by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ and see what happens:

\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} ad - bc & 0 \\ 0 & ad - bc \end{bmatrix} = (ad - bc)I.

If the number $ad - bc$ is not zero, dividing by it gives the inverse. The number $ad - bc$ is the determinant of $A$ , written $\det A$ . The next unit studies it in depth.

Inverse of a 2 × 2 matrix

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $ad - bc \ne 0$ , then

A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

If $ad - bc = 0$ , then $A$ is not invertible.

In words: swap the diagonal entries, negate the off-diagonal entries and divide by the determinant.

Solving systems with an inverse

If $A$ is invertible, then for every $\mathbf{b}$ in $\mathbb{R}^n$ the equation $A\mathbf{x} = \mathbf{b}$ has exactly one solution, namely $\mathbf{x} = A^{-1}\mathbf{b}$ . It is a solution because $A(A^{-1}\mathbf{b}) = I\mathbf{b} = \mathbf{b}$ . It is the only one because any solution $\mathbf{u}$ satisfies $\mathbf{u} = A^{-1}A\mathbf{u} = A^{-1}\mathbf{b}$ .

Worked example: Using the formula

Solve $\begin{aligned} 3x_1 + 5x_2 &= 4 \\ x_1 + 2x_2 &= 1 \end{aligned}$ using an inverse.

The coefficient matrix is $A = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$ , and $\det A = 3(2) - 5(1) = 1$ . So

A^{-1} = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}, \qquad \mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}.

Check: $3(3) + 5(-1) = 4$ and $3 + 2(-1) = 1$ .

In practice, row reducing $\begin{bmatrix} A & \mathbf{b} \end{bmatrix}$ is faster than computing $A^{-1}$ for a single system. The inverse earns its keep in theory, and when you must solve with the same $A$ for many different right-hand sides.

Algebra of inverses

If $A$ and $B$ are invertible $n \times n$ matrices, then:

$A^{-1}$ is invertible and $(A^{-1})^{-1} = A$ .
$AB$ is invertible and $(AB)^{-1} = B^{-1}A^{-1}$ .
$A^T$ is invertible and $(A^T)^{-1} = (A^{-1})^T$ .

For rule 2, check directly: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$ , and similarly in the other order. The order reverses for the same reason you take off your shoes before your socks: the last thing done is the first thing undone.

Common mistake

$(AB)^{-1}$ is $B^{-1}A^{-1}$ , not $A^{-1}B^{-1}$ . Since matrix multiplication doesn't commute, these are usually different. Also, $(A + B)^{-1}$ has no simple formula; it need not even exist when $A$ and $B$ are both invertible (try $B = -A$ ).

Elementary matrices

An elementary matrix is what you get by performing one elementary row operation on $I_n$ . The key fact: performing that row operation on any $n \times m$ matrix $A$ gives the same result as computing $EA$ .

For instance, with $E = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}$ (add $-3$ times row 1 to row 3), $EA$ is $A$ with that replacement carried out. Every row operation can be undone by another row operation of the same type, so every elementary matrix is invertible, and $E^{-1}$ is the elementary matrix of the reverse operation. Here, $E^{-1}$ adds $+3$ times row 1 to row 3.

Now suppose a sequence of row operations reduces $A$ to $I$ : $E_k \cdots E_2E_1A = I$ . Then $E_k \cdots E_1$ is exactly $A^{-1}$ . Applying those same operations to $I$ produces $E_k \cdots E_1 I = A^{-1}$ . That gives an algorithm.

Finding A⁻¹ by row reduction

Row reduce the augmented matrix $\begin{bmatrix} A & I \end{bmatrix}$ .

If $A$ is row equivalent to $I$ , then $\begin{bmatrix} A & I \end{bmatrix}$ reduces to $\begin{bmatrix} I & A^{-1} \end{bmatrix}$ .
Otherwise, $A$ is not invertible.

Worked example: A 3 × 3 inverse

Find the inverse of $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ , if it exists.

Start with $\begin{bmatrix} A & I \end{bmatrix}$ and subtract row 1 from row 3, then row 2 from row 3:

\left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] \sim \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & -1 & -1 & 1 \end{array}\right].

Multiply row 3 by $-1$ , then subtract row 3 from rows 1 and 2:

\sim \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 & -1 \end{array}\right] \sim \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -1 & 1 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 & -1 \end{array}\right].

The left block is $I$ , so

A^{-1} = \begin{bmatrix} 0 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & -1 \end{bmatrix}.

Check one column: $A\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 + 0 + 1 \\ 0 - 1 + 1 \\ 0 - 1 + 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ , the first column of $I$ .

Worked example: A singular matrix

Is $B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 3 & 4 \end{bmatrix}$ invertible?

Subtract $2$ times row 1 from row 2 and row 1 from row 3:

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}.

There is a row of zeros, so $B$ has only two pivots and cannot be row reduced to $I$ . $B$ is not invertible. (You can see why: row 3 of $B$ equals row 2 minus row 1.)

Tip

Always check a computed inverse by multiplying it by $A$ , at least for one column. An arithmetic slip anywhere in the row reduction shows up immediately.

Practice

Practice 1

Let $A = \begin{bmatrix} 4 & 3 \\ 3 & 2 \end{bmatrix}$ . What is the $(2, 2)$ entry of $A^{-1}$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

For what value of $k$ is $\begin{bmatrix} 2 & k \\ 6 & 9 \end{bmatrix}$ not invertible?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Use an inverse to solve $\begin{aligned} 2x_1 + x_2 &= 1 \\ 5x_1 + 3x_2 &= -2. \end{aligned}$ Enter $(x_1, x_2)$ .

Enter a point like (2, -3)

Practice 4

An invertible matrix $A$ has $A^{-1} = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ . Solve $A\mathbf{x} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ . Enter $\mathbf{x}$ as $(x_1, x_2)$ .

Enter a point like (2, -3)

Practice 5

$A$ , $B$ and $C$ are invertible $n \times n$ matrices. Which expression equals $(ABC)^{-1}$ ?

Practice 6

Let $L = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 4 & 1 \end{bmatrix}$ . Find the entry in row 3, column 1 of $L^{-1}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$A$ is an invertible $3 \times 3$ matrix with $A^2 = 2A$ . What is the sum of the diagonal entries of $A$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.