Let $$A \in \mathbb{F}^{n \times n}$$. How do we find the inverse of $$A$$, if it exists?

Before we look at a general procedure, let us look at the matrix inverses of elementary matrices.

Recall that if $$E$$ is an $$m\times m$$ elementary matrix and $$B$$ is an $$m\times n$$ matrix, then $$E B$$ is the matrix obtained from $$B$$ by performing the elementary row operation to which the matrix $$E$$ corresponds.

What should $$E^{-1}$$ be? Clearly, it needs to be a matrix that undoes what $$E$$ does to $$B$$.

If $$E$$ corresponds to the elementary row operation $$R_i \leftarrow R_i + \alpha R_j$$ where $$i \neq j$$ and $$\alpha \neq 0$$, then $$E^{-1}$$ corresponds to the elementary row operation $$R_i \leftarrow R_i - \alpha R_j$$. For example, the inverse of $$\begin{bmatrix} 1 & 0 \\ 7 & 1\end{bmatrix}$$ is $$\begin{bmatrix} 1 & 0 \\ -7 & 1\end{bmatrix}$$.

If $$E$$ corresponds to the elementary row operation $$R_i \leftrightarrow R_j$$ where $$i \neq j$$, then $$E$$ is its own inverse because performing $$R_i \leftrightarrow R_j$$ twice in a row in effect does nothing. For example, the matrix $$E=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}$$ is its own inverse.

If $$E$$ corresponds to the elementary row operation $$R_i \leftarrow \alpha R_i$$ where $$\alpha \neq 0$$, then $$E^{-1}$$ corresponds to the elementary row operation $$R_i \leftarrow \alpha^{-1} R_i$$. For example, the inverse of $$\begin{bmatrix} 1 & 0 \\ 0 & -3\end{bmatrix}$$ is $$\begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{3}\end{bmatrix}$$.

We now describe a way of obtaining $$A^{-1}$$, if it exists, using row reduction.

Note that we are seeking a matrix $$B \in \mathbb{F}^{n\times n}$$ such that $$AB = I_n$$. Comparing the $$j$$th columns on both sides, we have $$AB_j = e_j$$ for all $$j = 1,\ldots,n$$ where $$e_j$$ denote the $$j$$th column of $$I_n$$. In other words, we can find $$B$$ by solving $$n$$ systems of linear equations having the same coefficient matrix.

Hence, we can find the inverse matrix of $$A$$ by forming the matrix $$[A\mid I_n]$$ and transforming it to reduced row-echelon form via row reduction. If the resulting matrix has the form $$[I_n \mid B]$$, then $$B = A^{-1}$$.

As an illustration, consider $$A = \begin{bmatrix} 1 & 2 \\ 3 & -1\end{bmatrix}$$. To find the inverse of $$A$$, we want to find a matrix $$B = \begin{bmatrix} x_1 & y_1 \\ x_2 & y_2 \end{bmatrix}$$ such that $$A B = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$.

Letting $$C = AB$$, we see that the first column of $$C$$ is $$A \begin{bmatrix} x_1 \\x_2\end{bmatrix}$$ and the second column of $$C$$ is $$A \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$$. It follows that $A B = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ is equivalent to $A \begin{bmatrix} x_1 \\x_2\end{bmatrix} = \begin{bmatrix} 1 \\ 0\end{bmatrix} \text{ and } A \begin{bmatrix} y_1 \\y_2\end{bmatrix} = \begin{bmatrix} 0 \\ 1\end{bmatrix}.$ Hence, we can obtain $$B$$ by solving the above two systems. To solve both systems in one stroke, we set up the extended augmented matrix $[A \mid I_2] = \left[\begin{array}{cc|cc} 1 & 2 & 1 & 0\\ 3 & -1 & 0 & 1\end{array}\right].$ Row-reducing this extended augmented matrix gives $\left[\begin{array}{cc|cc} 1 & 0 & \frac{1}{7} & \frac{2}{7}\\ 0 & 1 & \frac{3}{7} & -\frac{1}{7}\end{array}\right].$ As the first two columns of this matrix form the identity matrix, we obtain that $$A^{-1} = \begin{bmatrix} \frac{1}{7} & \frac{2}{7}\\ \frac{3}{7} & -\frac{1}{7}\end{bmatrix}.$$

### Example

Let $$A\in\mathbb{C}^{2\times 2}$$ be given by $$\begin{bmatrix} 1 & i \\ 0 & 2\end{bmatrix}$$. To compute $$A^{-1}$$, we row reduce $$\left[\begin{array}{cc|cc} 1 & i & 1 & 0\\ 0 & 2 & 0 & 1\end{array}\right]$$ as follows:

$\left[\begin{array}{cc|cc} 1 & i & 1 & 0\\ 0 & 2 & 0 & 1\end{array}\right] \stackrel{R_2\leftarrow \frac{1}{2}R_2}{\longrightarrow} \left[\begin{array}{cc|cc} 1 & i & 1 & 0\\ 0 & 1 & 0 & \frac{1}{2}\end{array}\right] \stackrel{R_1\leftarrow R_1+(-i)R_2}{\longrightarrow} \left[\begin{array}{cc|cc} 1 & 0 & 1 & -\frac{1}{2}i\\ 0 & 1 & 0 & \frac{1}{2}\end{array}\right]$ Hence, $$A^{-1}=\begin{bmatrix} 1 & -\frac{1}{2}i \\ 0 & \frac{1}{2} \end{bmatrix}$$.

## Exercises

1. For each of the following matrices, determine if it has an inverse. If it does, give it.

1. $$\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$

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

3. $$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

4. $$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$$

2. Find all values of $$a$$ such that the matrix $$\begin{bmatrix} 1 & 0 & 2\\ -1 & 1 & a \\ 0 & a & -1\end{bmatrix}$$ is invertible.