Invertible matrices

A square matrix $A$ is invertible if there exists a matrix $B$ such that $AB = BA = I$. When it exists, $B$ is unique and is denoted $A^{-1}$. Inverses generalise the reciprocal $1/a$ from numbers. They let us solve matrix equations $AX = C$ by writing $X = A^{-1} C$, just as we would solve $ax = c$ by $x = c/a$ in ordinary algebra. Not every nonzero matrix has an inverse , only those whose determinant is nonzero, a fact developed in the next chapter.

Definition

Let $A$ be a square matrix of order $n$. $A$ is invertible (or non-singular) if there exists an $n \times n$ matrix $B$ with $AB = BA = I_n$. The matrix $B$ is the inverse and is denoted $A^{-1}$.

If no such $B$ exists, $A$ is called singular.

Uniqueness of the inverse

Theorem. If $A$ has an inverse, it is unique.

Proof. Suppose $B$ and $C$ both satisfy $AB = BA = I$ and $AC = CA = I$. Then $B = BI = B(AC) = (BA)C = IC = C$. $\blacksquare$

Properties of inverses

For invertible $A$ and $B$ of the same order:

1. $(A^{-1})^{-1} = A$.
2. $(AB)^{-1} = B^{-1} A^{-1}$. Reverse the order.
3. $(A^T)^{-1} = (A^{-1})^T$.
4. $(\lambda A)^{-1} = \tfrac{1}{\lambda} A^{-1}$ for nonzero scalar $\lambda$.

Proof of (2). $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$. Similarly $(B^{-1}A^{-1})(AB) = I$. $\blacksquare$

Inverse by elementary row operations

To find $A^{-1}$, write $[A | I]$ as a single augmented matrix and apply elementary row operations until the left half becomes $I$. Whatever remains on the right is $A^{-1}$. The reason: row operations are left-multiplication by elementary matrices; if their product turns $A$ into $I$, the same product turns $I$ into $A^{-1}$.

If at any stage a zero row appears on the left, $A$ is singular and has no inverse.

Worked computation

Find the inverse of $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

Set up: $[A | I] = \begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 3 & 4 & | & 0 & 1 \end{pmatrix}$.

$R_2 \to R_2 - 3 R_1$: $\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & -2 & | & -3 & 1 \end{pmatrix}$.

$R_2 \to -\tfrac{1}{2} R_2$: $\begin{pmatrix} 1 & 2 & | & 1 & 0 \\ 0 & 1 & | & 3/2 & -1/2 \end{pmatrix}$.

$R_1 \to R_1 - 2 R_2$: $\begin{pmatrix} 1 & 0 & | & -2 & 1 \\ 0 & 1 & | & 3/2 & -1/2 \end{pmatrix}$.

So $A^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$. Verify: $A A^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. $\checkmark$

Quick $2 \times 2$ formula

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $ad - bc \neq 0$,

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

The number $ad - bc$ is the determinant of $A$. If $ad - bc = 0$, $A$ is singular.

Solving linear systems via inverses

If $AX = B$ and $A$ is invertible, then $X = A^{-1} B$. This is the matrix method of solving a linear system. It is computationally less efficient than direct row-reduction, but it has theoretical importance , it expresses the solution in closed form.

Worked examples

Example 1. Find the inverse of $A = \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix}$.

$ad - bc = 8 - 7 = 1$. $A^{-1} = \begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix}$.

Example 2. Find the inverse of $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix}$ by row operations.

$R_2 \to R_2 - 4 R_3$ in $[A|I]$:

After more steps, $A^{-1} = \begin{pmatrix} 1 & -2 & 5 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{pmatrix}$.

Example 3. Show that $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ is singular.

$\det = 4 - 4 = 0$. No inverse.

Example 4. Solve $AX = B$ where $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}, B = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$.

$A^{-1} = \tfrac{1}{1}\begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$. $X = A^{-1}B = \begin{pmatrix} 10 - 9 \\ -5 + 6 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Example 5. Verify $(AB)^{-1} = B^{-1}A^{-1}$ for $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.

$AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, $\det = 1$, $(AB)^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.

$A^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$. $B^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$. Product $B^{-1}A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$. $\checkmark$

Example 6. Show that an elementary matrix is always invertible. Find the inverse of the row-swap matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

$\det = -1 \neq 0$. The inverse is itself: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = I$. So the matrix is involutory.

Try it yourself

1. Find the inverse of $\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$.
2. Find the inverse of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
3. Is $\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$ invertible?
4. Find $A^{-1}$ for $A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
5. Verify $(AB)^{-1} = B^{-1}A^{-1}$ for matrices of your choice.
6. Solve $\begin{cases} 2x + 3y = 5 \\ x + 2y = 3 \end{cases}$ using inverses.
7. Show $A^{-1} = A^T$ for $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.
8. Find a $2 \times 2$ invertible matrix that is its own inverse.
9. Show: if $A^2 = A$ and $A$ is invertible, then $A = I$.
10. Show: $(\lambda A)^{-1} = (1/\lambda) A^{-1}$ for nonzero $\lambda$.
11. Find the inverse of $A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix}$.
12. If $A$ is invertible and $A^3 = O$, derive a contradiction.
13. Find the inverse of $A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}$.
14. If $AB = I$ for square $A, B$ of the same order, must $BA = I$? (Answer: yes , explain.)

Pitfalls / Tricks

• Always check that the matrix is square before asking for an inverse.
• Always verify $\det A \neq 0$ , a zero determinant means singular.
• The reversal $(AB)^{-1} = B^{-1}A^{-1}$ is crucial; do not forget it.
• For $2 \times 2$ matrices the explicit formula is fastest; for larger matrices row operations beat hand-cofactor methods.
• The inverse of an orthogonal matrix (like a rotation) is its transpose.

This concludes the chapter on matrices. The determinant , central to detecting singularity , is the subject of Chapter 4.