Inverse Matrices

The inverse of an inverse matrix is the original matrix:

\begin{align} (A^{-1})^{-1} = A \end{align}

The inverse of two matrices multiplied by each other is:

\begin{align} (AB)^{-1} = B^{-1}A^{-1} \end{align}

Given a \(2 \times 2\) matrix \(A\) such that:

\begin{align} A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \notag \end{align}

Call \(ad-bc\) the determinant of \(A\). If \(ad-bc \neq 0\), then:

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

Note that if \(ad=bc\), then the matrix is not invertible.

Consider the elementary matrix \(E\): a matrix obtained from \(I\) by performing a single row operation. Notice that applying that single row operation to another matrix \(A\) is equivalent to computing \(EA\).

If \(A\) is invertible, then its reduced row echelon form is \(I\). This is because if \(A\) is invertible, then there is a unique solution to \(A\mathbf{x}=\mathbf{b}\) (which is \(A^{-1}\mathbf{b}\)), which means that there must be a pivot in every column of the RREF of \(A\). If there is a pivot in every column of \(A\), then by the definition of RREF those pivots must be 1, and every other number is 0, which becomes the identity matrix.

Therefore, if \(A\) is invertible, then we can row reduce \(A\) to \(I\) using row operations. We can encode these row operations using elementary matrices like so:

\begin{align} E_kE_{k-1}\cdots E_1A &= I \notag \end{align}

Thus, it can be seen that in general, the inverse of \(A\) is:

\begin{align} A^{-1} = E_kE_{k-1}\cdots E_1 \end{align}

To find this series of elementary matrices, we can create the augmented matrix \(\begin{bmatrix} A & I \end{bmatrix}\). When we row reduce this augmented matrix such that the \(A\) part becomes \(I\), the \(I\) part of the augmented matrix will become \(A^{-1}\) (the series of row operations encoded as the product of elementary matrices).

We can use Cramer's rule to find inverses of matrices as well. We want to find the inverse of a square matrix \(A\) such that \(A\cdot A^{-1}=I\).

Let the first column of \(A^{-1}\) be:

\begin{align} \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \notag \end{align}

Then, by matrix multiplication of the columns, we see that:

\begin{align} A\mathbf{x} = e_1 \notag \end{align}

We can then find each entry in \(\mathbf{x}\) using Cramer's rule. Without loss of generality, consider \(x_3\):

\begin{align} x_3 = \frac{\det[A_3(e_1)]}{\det A} \notag \end{align}

Doing cofactor expansion in the third column of \(A_3(e_1)\) shows us that \(\det[A_3(e_1)]\) is \(C_{13}\), the cofactor of the entry corresponding to the first row and third column. Therefore,

\begin{align} x_3 = \frac{C_{13}}{\det A} \notag \end{align}

Similarly, for the others, we get:

\begin{align} A^{-1} = \frac{1}{\det A} \begin{bmatrix} C_{11} & C_{21} & \cdots & C_{n1} \\ C_{12} & \ddots \\ \vdots \\ C_{1n} & & & C_{nn} \end{bmatrix} \notag \end{align}

The matrix above in the equation is called \(\text{adj }A\), the adjugate matrix of \(A\). This is where you replace every entry of \(A\) by its cofactor and then take its transpose. Thus, we can simplify this formula to:

\begin{align} \boxed{A^{-1} = \frac{\text{adj }A}{\det A}} \end{align}

This can also be used to derive the base formula for the inverse of a \(2 \times 2\) matrix.