# Inverse of a Matrix

Let $A$ be an $n\times n$ matrix with $\det A\ne 0$. (A square matrix whose determinant is not equal to $0$ is called non-singular.) Let $X=(x_{ij})$ be an unknown $n\times n$ matrix such that
$AX=I$. Then
$$x_{1j}A^1+\cdots+x_{nj}A^n=E^j.$$
This is a system of linear equations and as we studied here, it can be solved by Cramer’s Rule as
\begin{align*}
x_{ij}&=\frac{\det(A^1,\cdots,E^j,\cdots A^n)}{\det A}\\
&=\frac{1}{\det A}\left|\begin{array}{ccccc}
a_{11} & \cdots & 0 & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots\\
a_{j1} & \cdots & 1 & \cdots & a_{jn}\\
\vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & 0 & \cdots & a_{nn}
\end{array}\right|\\
&=\frac{1}{\det A}(-1)^{i+j}\det(A_{ji})
\end{align*}
for $i=1,\cdots,n$. If we show that $XA=I$ as well, then $X$ would be the inverse $A^{-1}$ and
$$A^{-1}=\frac{1}{\det A}{}^t((-1)^{i+j}\det(A_{ij})).$$
$\det({}^tA)=\det A\ne 0$, so we can find an $n\times n$ matrix $Y$ such that ${}^tAY=I$. Taking transposes, we obtain ${}^tYA=I$. Now,
$$I={}^tYA={}^tYIA={}^tY(AX)A={}^tYA(XA)=XA.$$

Example. Let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ with $\det A=ad-bc\ne 0$. Then $A^{-1}$ is given by
$$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}.$$

Example. Find the inverse of the matrix
$$A=\begin{pmatrix} 3 & 1 & -2\\ -1 & 1 & 2\\ 1 & -2 & 1 \end{pmatrix}.$$

Solution. $\det A=16$. We find
\begin{align*}
A_{11}=\begin{pmatrix}
1 & 2\\
-2 & 1
\end{pmatrix},\ A_{12}=\begin{pmatrix}
-1 & 2\\
1 & 1
\end{pmatrix},\ A_{13}=\begin{pmatrix}
-1 & 1\\
1 & -2
\end{pmatrix},\\
A_{21}=\begin{pmatrix}
1 & -2\\
-2 & 1
\end{pmatrix},\ A_{22}=\begin{pmatrix}
3 & -2\\
1 & 1
\end{pmatrix},\ A_{23}=\begin{pmatrix}
3 & 1\\
1 & -2
\end{pmatrix},\\
A_{31}=\begin{pmatrix}
1 & -2\\
1 & 2
\end{pmatrix},\ A_{32}=\begin{pmatrix}
3 & -2\\
-1 & 2
\end{pmatrix},\ A_{33}=\begin{pmatrix}
3 & 1\\
-1 & 1
\end{pmatrix}.
\end{align*}
Hence by the formula we obtain
\begin{align*}
A^{-1}&=\frac{1}{16}{}^t\begin{pmatrix}
\det(A_{11}) & -\det(A_{12}) & \det(A_{13})\\
-\det(A_{21}) & \det(A_{22}) & -\det(A_{23})\\
\det(A_{31}) &-\det(A_{32}) & \det(A_{33})
\end{pmatrix}\\
&=\frac{1}{16}{}^t\begin{pmatrix}
5 & 3 & 1\\
3 & 5 & 7\\
4 & -4 & 4
\end{pmatrix}\\
&=\frac{1}{16}\begin{pmatrix}
5 & 3 & 4\\
3 & 5 &-4\\
1 & 7 & 4
\end{pmatrix}.
\end{align*}

We introduce the following theorem without proof:

Theorem. For any two $n\times n$ matrices $A$, $B$,
$$\det(AB)=\det A\det B.$$

As a special case, we obtain:

Corollary. For an invertible matrix $A$,
$$\det(A^{-1})=\frac{1}{\det A}.$$

Proof. $AA^{-1}=I$, so by the theorem
$$1=\det(AA^{-1})=\det A\det (A^{-1}).$$
Thus proving the formula for the inverse.

This site uses Akismet to reduce spam. Learn how your comment data is processed.