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 \(A^{-1} = \begin{bmatrix} \frac{1}{7} & \frac{2}{7}\\ \frac{3}{7} & -\frac{1}{7}\end{bmatrix}.\)
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, \(A^{-1}=\begin{bmatrix} 1 & -\frac{1}{2}i \\ 0 & \frac{1}{2} \end{bmatrix}\).
For each of the following matrices, determine if it has an inverse. If it does, give it.
\(\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\)
Find all values of \(a\) such that the matrix \(\begin{bmatrix} 1 & 0 & 2\\ -1 & 1 & a \\ 0 & a & -1\end{bmatrix}\) is invertible.