Linear Algebra · Book or PDF

A First Course in Linear Algebra

A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 5 · pp. 54–63

Chapter 5: Matrix Inverses

Some square matrices can be 'undone' by another matrix, just as dividing undoes multiplying. This chapter defines the inverse of a matrix, shows how to find it with row operations, and explains why the method works. It also covers a shortcut for $2\times 2$ matrices and the rule for the inverse of a product.

What an inverse is

A square matrix $A$ is invertible (also called non-singular) if there is a matrix $B$ of the same size with $AB = I$ and $BA = I$ . We write $B = A^{-1}$ . A square matrix with no inverse is singular. Only square matrices can be invertible. (p. 54)

Example: a matrix that swaps two rows undoes itself when applied twice. So it is its own inverse. A matrix with a row of zeros is singular. (p. 54)

Two facts help here:

If an inverse exists, it is unique. That is why we can say 'the' inverse. (p. 54)
Some matrices are always invertible: elementary matrices, orthogonal matrices, and diagonal matrices with no zeros on the main diagonal. For a diagonal matrix with nonzero entries $a, b, c$ , the inverse is diagonal with entries $1/a, 1/b, 1/c$ . (p. 54)

Not every matrix has an inverse, so always ask whether one exists.

Finding an inverse by row operations

The method extends Gaussian elimination (GE) and works for square matrices of any size. (pp. 55–56, 58)

Write $A$ next to an identity matrix of the same size.
Use row operations to get 1s on the main diagonal and 0s below it. This is the elimination stage.
Keep going to clear the entries above the diagonal. Work from the last column back to the first.
Apply every operation to both sides. When the left side becomes $I$ , the right side is $A^{-1}$ .

Worked example. Let $A$ have rows $(1,1,1)$ , $(1,2,3)$ , $(0,1,1)$ . Elimination gives an upper triangular matrix with rows $(1,1,1)$ , $(0,1,2)$ , $(0,0,1)$ . Continuing upward gives

$A^{-1} = \begin{bmatrix} 1 & 0 & -1\\ 1 & -1 & 2\\ -1 & 1 & -1\end{bmatrix}.$

You can check this by multiplying. The first row of $A$ times the columns of this matrix gives $1, 0, 0$ , and the other rows work out the same way. (pp. 55, 58)

A check is always worth doing: multiply $A$ by your answer $B$ . If you do not get $I$ , you made a mistake. (p. 58)

Why the method works

Each row operation is the same as multiplying by an elementary matrix. Suppose the operations that reduce $A$ to $I$ correspond to $E_1, E_2, \ldots, E_s$ . Then

$E_s\cdots E_2E_1A = I.$

Let $B = E_s\cdots E_1$ . Then $BA = I$ . (p. 56)

We also need $AB = I$ . Each $E_i$ has an inverse $F_i$ . Multiplying the equation above on the left by $F_1F_2\cdots F_s$ gives $A = F_1\cdots F_s$ . Then

$AB = F_1\cdots F_sE_s\cdots E_1 = I,$

because the inner pairs $F_sE_s$ , $F_{s-1}E_{s-1}$ , and so on, each cancel to $I$ . So $B$ is the inverse of $A$ . (p. 56)

Now look at what happens to the identity matrix. Applying the same operations to $I$ produces $E_s\cdots E_1I = B = A^{-1}$ . That is why the right-hand side ends up as the inverse. (p. 56)

When there is no inverse

A square matrix is invertible exactly when elimination reaches a form with 1s on the main diagonal and 0s below it. If a row of zeros appears, the process stops and the matrix is singular. (pp. 58–60)

Example. Take rows $(1,2,3)$ , $(1,1,2)$ , $(0,1,1)$ . Elimination ends with the last row $(0,0,0)$ . There is no way to put a 1 in the bottom corner, so there is no inverse. (pp. 58–59)

Why a zero row rules out an inverse. First, a useful fact: if a $p\times p$ matrix $A$ has a last row of all zeros, then $AX$ also has a last row of zeros for any $p\times p$ matrix $X$ . The entries in that row are sums of terms $a_{pk}x_{kj}$ , and every $a_{pk}$ is 0. So each sum is 0. (p. 59)

Now suppose a matrix with a zero last row were invertible. Multiplying it by its inverse would have to give $I$ . But the product would have a zero last row, and $I$ does not. This is a contradiction, so no inverse exists. (pp. 59–60)

The $2\times 2$ formula

For a $2\times 2$ matrix, if $ad-bc\neq 0$ :

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

To use it, swap the diagonal entries, change the signs of the other two, and divide by $ad-bc$ . (p. 59)

Example. For $\begin{bmatrix}1&2\\1&3\end{bmatrix}$ we have $ad-bc = 3-2 = 1$ . The inverse is $\begin{bmatrix}3&-2\\-1&1\end{bmatrix}$ . Multiplying the two matrices gives $I$ . (pp. 59–60)

If $ad-bc = 0$ , the formula cannot be used and the matrix is singular.

Checking an inverse with only one product

The definition asks for both $AB = I$ and $BA = I$ . For square matrices, one is enough. The other follows automatically. (pp. 58–60)

Here is the reasoning. Suppose $AB = I$ but $A$ were singular. Then some invertible matrix $X$ makes $XA$ have a zero last row. So $XAB$ would have a zero last row too. But $XAB = XI = X$ , and an invertible matrix cannot have a zero last row. This contradiction shows $A$ is invertible. Multiplying $AB = I$ on the left by $A^{-1}$ then gives $B = A^{-1}$ . (p. 59)

In practice you only need to compute one product to verify an answer.

Inverse of a product

If $A$ and $B$ are invertible matrices of the same size, then $AB$ is invertible and

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

Check: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$ . Multiplying in the other order gives $I$ as well. Since inverses are unique, this is the inverse. (p. 60)

The order reverses, like taking off socks and shoes: the shoes (last on) come off first. For three matrices, $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ . (p. 60)

Common mistake. Many students write $(AB)^{-1} = A^{-1}B^{-1}$ . Matrix multiplication is generally not commutative, so this fails. The factors $B$ and $B^{-1}$ must meet in the middle and cancel first. (p. 60)

The same reversal appears with elementary matrices. The inverse of $E_r\cdots E_1$ is $F_1\cdots F_r$ , where each $F_i$ is the inverse of $E_i$ . (p. 60)

Check yourself

Why can a $2\times 3$ matrix never be invertible?
Find the inverse of $\begin{bmatrix}2&1\\5&3\end{bmatrix}$ using the $2\times 2$ formula, then verify it with one product.
During row reduction of a $3\times 3$ matrix, the last row becomes all zeros. What does this tell you, and why?
If $A$ and $B$ are invertible and the same size, write $(AB)^{-1}$ and explain why $A^{-1}B^{-1}$ is generally wrong.
What is the inverse of the diagonal matrix with diagonal entries $2, -4, 5$ ?

A First Course in Linear Algebra

Chapter 5: Matrix Inverses

What an inverse is

Finding an inverse by row operations

Why the method works

When there is no inverse

The 2×22\times 22×2 formula

Checking an inverse with only one product

Inverse of a product

Check yourself

Chapter 5: Matrix Inverses

What an inverse is

Finding an inverse by row operations

Why the method works

When there is no inverse

The 2×22\times 22×2 formula

Checking an inverse with only one product

Inverse of a product

Check yourself

The $2\times 2$ formula

The $2\times 2$ formula