Adjoint and inverse of a matrix

For num­bers, divid­ing by $a$ is the same as mul­ti­ply­ing by $\displaystyle \frac{1}{a}$, the num­ber with $\displaystyle a \times \frac{1}{a} = 1$. Matri­ces have no divi­sion, but many square matri­ces have an inverse: a matrix $A^{-1}$ with $AA^{-1} = A^{-1}A = I$. The inverse is what lets you "divide" a matrix equa­tion, and it is the key to solv­ing sys­tems of equa­tions in the next les­son. This les­son shows how to find the inverse using the adjoint, and when an inverse exists.

The adjoint#

The adjoint of a square matrix $A$, writ­ten $\text{adj}\,A$, is the trans­pose of the matrix of cofac­tors of $A$.

So to find $\text{adj}\,A$: replace each entry of $A$ by its cofac­tor, then trans­pose.

Short­cut for $2 \times 2$. For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the cofac­tors are $C_{11} = d$, $C_{12} = -c$, $C_{21} = -b$, $C_{22} = a$. Trans­pos­ing gives

$$\text{adj}\,A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

In words: swap the two diag­o­nal entries and change the signs of the other two.

The key fact#

Result. For any square matrix $A$,

$$A(\text{adj}\,A) = (\text{adj}\,A)A = |A|\,I$$

The rea­son: the diag­o­nal entries of $A(\text{adj}\,A)$ are rows of $A$ times their own cofac­tors, which is the expan­sion of $|A|$. The off-diag­o­nal entries are rows times the cofac­tors of a dif­fer­ent row, which is the expan­sion of a deter­mi­nant with two equal rows, and so is $0$.

The inverse#

If $|A| \neq 0$, divide the key fact by $|A|$:

$$\displaystyle A\left(\frac{1}{|A|}\text{adj}\,A\right) = I$$

So

$$\displaystyle A^{-1} = \frac{1}{|A|}\,\text{adj}\,A$$

A square matrix has an inverse exactly when it is non-sin­gu­lar, that is, $|A| \neq 0$. A sin­gu­lar matrix ($|A| = 0$) has no inverse.

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

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

Exam­ple 1#

Find the inverse of $A = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$ and check it.

$|A| = 6 - 5 = 1$, so $A^{-1} = \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}$.

Check: $AA^{-1} = \begin{bmatrix} 6 - 5 & -15 + 15 \\ 2 - 2 & -5 + 6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.

Exam­ple 2#

Find the inverse of $B = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}$.

$|B| = 24 - 14 = 10$, so

$$\displaystyle B^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{5} & -\frac{7}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}$$

Exam­ple 3#

Find the inverse of $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 0 & 0 \end{bmatrix}$.

Step 1: the deter­mi­nant. Expand along the third row, which has two zeros:

$$|A| = 1 \times \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = 2 - 3 = -1$$

$|A| \neq 0$, so the inverse exists.

Step 2: the cofac­tors.

$$C_{11} = \begin{vmatrix} 3 & 2 \\ 0 & 0 \end{vmatrix} = 0, \quad C_{12} = -\begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} = 2, \quad C_{13} = \begin{vmatrix} 1 & 3 \\ 1 & 0 \end{vmatrix} = -3$$

$$C_{21} = -\begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} = 0, \quad C_{22} = \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = -1, \quad C_{23} = -\begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = 1$$

$$C_{31} = \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = -1, \quad C_{32} = -\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = -3, \quad C_{33} = \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 5$$

Step 3: the adjoint. The cofac­tor matrix is $\begin{bmatrix} 0 & 2 & -3 \\ 0 & -1 & 1 \\ -1 & -3 & 5 \end{bmatrix}$; trans­pos­ing,

$$\text{adj}\,A = \begin{bmatrix} 0 & 0 & -1 \\ 2 & -1 & -3 \\ -3 & 1 & 5 \end{bmatrix}$$

Step 4: the inverse.

$$\displaystyle A^{-1} = \frac{1}{-1}\,\text{adj}\,A = \begin{bmatrix} 0 & 0 & 1 \\ -2 & 1 & 3 \\ 3 & -1 & -5 \end{bmatrix}$$

Check the first row of $A$ against each col­umn of $A^{-1}$: $(2)(0) + (1)(-2) + (1)(3) = 1$, $(2)(0) + (1)(1) + (1)(-1) = 0$, $(2)(1) + (1)(3) + (1)(-5) = 0$. That is the first row of $I$, as it should be.

Prop­er­ties of the inverse#

For non-sin­gu­lar matri­ces of the same order:

1. $(A^{-1})^{-1} = A$
2. $(AB)^{-1} = B^{-1}A^{-1}$ (the rever­sal law again)
3. $(A')^{-1} = (A^{-1})'$
4. $\displaystyle |A^{-1}| = \frac{1}{|A|}$

Exam­ple 4#

If $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$, show that $A^2 - 4A + I = O$, and use this to find $A^{-1}$.

$$A^2 = \begin{bmatrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}, \qquad A^2 - 4A + I = \begin{bmatrix} 7 - 8 + 1 & 12 - 12 \\ 4 - 4 & 7 - 8 + 1 \end{bmatrix} = O$$

Mul­ti­ply $A^2 - 4A + I = O$ by $A^{-1}$: $A - 4I + A^{-1} = O$, so

$$A^{-1} = 4I - A = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$

which agrees with the $2 \times 2$ short­cut, since $|A| = 1$.

Prac­tice#

1. Write the adjoint of $\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$ and find its inverse.
2. Find the inverse of $\begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$.
3. Does $\begin{bmatrix} 6 & 4 \\ 3 & 2 \end{bmatrix}$ have an inverse? Give a rea­son.
4. For which value of $k$ does $\begin{bmatrix} k & 3 \\ 4 & 6 \end{bmatrix}$ have no inverse?
5. Find the adjoint and the inverse of $\begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$.
6. Find the inverse of $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$.
7. If $|A| = 5$ for a $3 \times 3$ matrix $A$, find $|A^{-1}|$.
8. Ver­ify $(AB)^{-1} = B^{-1}A^{-1}$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.
9. If $A^2 - 5A + 2I = O$ for some square matrix $A$, express $A^{-1}$ in terms of $A$ and $I$.

Answers#