The inverse of an n x n matrix A (if it exists) is defined to be the matrix B so that BA = I and AB = I, where I is the identity matrix. If the inverse exists, A is called invertible and its inverse is denoted by A-1.
To determine if an n x n matrix A is invertible, we can use the determinant: An n x n matrix A is invertible if and only if det(A) ≠ 0.
For an invertible matrix A, the system of linear equations Ax = b has the unique solution x = A-1b
Computing Inverses:
For a 2 x 2 matrix \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) we have a formula we can use to find the inverse:
\(A^{-1}=\frac{1}{\det(A)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.\)
For a more general n x n invertible matrix A, we use the following steps:
- Augment A with the identity matrix (I) of the same size.
- Use row operations to take A to reduced row echelon form while performing those operations on the augmented matrix, i.e., including I as well. This should result in I in the place where A was.
- Once the original A becomes I, the matrix in the place of where I was is the inverse A-1.
Here is a visualization: \(\begin{bmatrix}A|I\end{bmatrix}\to\ldots\to\begin{bmatrix}I|A^{-1}\end{bmatrix}.\)
Useful Facts:
Let A and B be invertible matrices.
\((1)\quad I^{-1}=I\)
\((2)\quad(A^{-1})^{-1}=A\)
\( (3)\quad(AB)^{-1}=B^{-1}A^{-1}\)
\((4)\quad(A^k)^{-1}=(A^{-1})^k,\mathrm{where~}k\in\)
\((5)\quad(aA)^{-1}=\frac{1}{a}A^{-1},\mathrm{where~}a\neq0\text{ is a scalar} \)
\((6)\quad(A^T)^{-1}=(A^{-1})^T,\mathrm{where~}T\text{ denotes the transpose of a matrix}\)
Example. Find the inverses of the following matrices.
\(\mathrm{(a)}\quad A=\begin{bmatrix}8&7\\1&2\end{bmatrix}\quad\mathrm{(b)}\quad A=\begin{bmatrix}3&7&2\\5&2&6\\1&0&4\end{bmatrix}\)
Solution Part (a).
First, we’ll find out if it is invertible by finding the determinant.
det(A) = (8)(2) - (1)(7) = 9 ≠ 0
Since the determinant is non-zero, we use the formula for a 2 X 2 matrix:
\(A^{-1}=\frac{1}{\det(A)}{\begin{bmatrix}2&-7\\-1&8\end{bmatrix}}=\frac{1}{9}{\begin{bmatrix}2&-7\\-3&8\end{bmatrix}}=\begin{bmatrix}\frac{2}{9}&\frac{-7}{9}\\\frac{-3}{9}&\frac{8}{9}\end{bmatrix}\)
Solution Part (b).
Again, find if A is invertible first by computing its determinant:
det(A) = 1(42 − 4) − 0 + 4(6 − 35)
= 38+ 4(−29)
= −78 ≠ 0
Apply the row reduction to A augmented by I:
\(R1\leftrightarrow R3\)
\(\left[\begin{array}{ccc|ccc}
3 & 7 & 2 & 1 & 0 & 0 \\
5 & 2 & 6 & 0 & 1 & 0 \\
1 & 0 & 4 & 0 & 0 & 1
\end{array}\right]\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 4 & 0 & 0 & 1 \\
5 & 2 & 6 & 0 & 1 & 0 \\
3 & 7 & 2 & 1 & 0 & 0
\end{array}\right]\)
\(R2\leftarrow R2-5R1\)
\(R3\leftarrow R3-3R1\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 4 & 0 & 0 & 1 \\
0 & 2 & -14 & 0 & 1 & -5 \\
0 & 7 & -10 & 1 & 0 & -3
\end{array}\right]\)
\(R2\leftarrow \frac{1}{2} R2\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 4 & 0 & 0 & 1 \\
0 & 1 & -7 & 0 & \frac{1}{2} & \frac{-5}{2} \\
0 & 7 & -10 & 1 & 0 & -3
\end{array}\right]\)
\(R3\leftarrow R3-7R2\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 4 & 0 & 0 & 1 \\
0 & 1 & -7 & 0 & \frac{1}{2} & \frac{-5}{2} \\
0 & 0 & 39 & 1 & \frac{-7}{2} & \frac{29}{2}
\end{array}\right]\)
\(R3\leftarrow \frac{1}{39} R3\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 4 & 0 & 0 & 1 \\
0 & 1 & -7 & 0 & \frac{1}{2} & \frac{-5}{2} \\
0 & 0 & 1 & \frac{1}{39} & \frac{-7}{78} & \frac{29}{78}
\end{array}\right]\)
\(R3\leftarrow R1-4R3\)
\(R2\leftarrow R2+7R3\)
\(\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & \frac{-4}{39} & \frac{14}{39} & \frac{-19}{39} \\
0 & 1 & 0 & \frac{7}{39} & \frac{-5}{39} & \frac{4}{39} \\
0 & 0 & 1 & \frac{1}{39} & \frac{-7}{78} & \frac{29}{78}
\end{array}\right]\)
Since we finally made it to the identity matrix on the left side, the right side is our inverse. We can factor out a 1/39 from all the entries to make it easier to read:
\(A^{-1}=\frac{1}{39}\left[\begin{array}{ccc}
-4 & 14 & -19 \\
7 & -5 & 4 \\
1 & \frac{-7}{2} & \frac{29}{2}
\end{array}\right]\)