# Find Inverse Matrices Using Adjoint Matrices

## Problem 546

Let $A$ be an $n\times n$ matrix.

The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be
$C_{ij}=(-1)^{ij}\det(M_{ij}),$ where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.

Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.

When $A$ is invertible, then its inverse can be obtained by the formula

$A^{-1}=\frac{1}{\det(A)}\Adj(A).$

For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.

(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$.

(b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$.

## Solution.

### (a) The Inverse Matrix of $A$.

Since $A$ is an upper triangular matrix, the determinant of $A$ is the product of diagonal entries.
Thus we have $\det(A)=-1\neq 0$, and hence $A$ is invertible.

To find the inverse using the formula, we first determine the cofactors $C_{ij}$ of $A$.
We have
\begin{align*}
C_{11}&=\begin{vmatrix}
-1 & 2\\
0& 1
0 & 2\\
0& 1
0 & -1\\
0& 0
\end{vmatrix}=0\6pt] C_{21}&=-\begin{vmatrix} 5 & 2\\ 0& 1 \end{vmatrix}=-5, \quad C_{22}=\begin{vmatrix} 1 & 2\\ 0& 1 \end{vmatrix}=1, \quad C_{23}=-\begin{vmatrix} 1 & 5\\ 0& 0 \end{vmatrix}=0 \\[6pt] C_{31}&=\begin{vmatrix} 5 & 2\\ -1& 2 \end{vmatrix}=12, \quad C_{32}=-\begin{vmatrix} 1 & 2\\ 0& 2 \end{vmatrix}=-2, \quad C_{33}=\begin{vmatrix} 1 & 5\\ 0& -1 \end{vmatrix}=-1. \end{align*} The the adjoint matrix of A is \begin{align*} \Adj(A)=C^{\trans}=\begin{bmatrix} -1 & -5 & 12 \\ 0 &1 &-2 \\ 0 & 0 & -1 \end{bmatrix}. \end{align*} Using the formula, we obtain the inverse matrix \[A^{-1}=\frac{1}{\det(A)}\Adj(A)=\begin{bmatrix} 1 & 5 & -12 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}.

### (b) The Inverse Matrix of $B$.

To check the invertibility of the matrix $B$, we compute the determinant of $B$.
The second column cofactor expansion yields that
\begin{align*}
\det(B)=\begin{vmatrix}
1 & 2\\
3& 1
\end{vmatrix}=-5 \neq 0.
\end{align*}
So the matrix $B$ is invertible.

Now the cofactors $C_{ij}$ of $B$ are
\begin{align*}
C_{11}&=\begin{vmatrix}
1 & 4\\
0& 1
0 & 4\\
3& 1
0 & 1\\
3& 0
\end{vmatrix}=-3 \6pt] C_{21}&=-\begin{vmatrix} 0 & 2\\ 0& 1 \end{vmatrix}=0, \quad C_{22}=\begin{vmatrix} 1 & 2\\ 3& 1 \end{vmatrix}=-5, \quad C_{23}=-\begin{vmatrix} 1 & 0\\ 3& 0 \end{vmatrix}=0 \\[6pt] C_{31}&=\begin{vmatrix} 0 & 2\\ 1& 4 \end{vmatrix}=-2, \quad C_{32}=-\begin{vmatrix} 1 & 2\\ 0& 4 \end{vmatrix}=-4, \quad C_{33}=\begin{vmatrix} 1 & 0\\ 0& 1 \end{vmatrix}=1. \end{align*} Hence the adjoint matrix of B is \[\Adj(B)=C^{\trans}=\begin{bmatrix} 1 & 0 & -2 \\ 12 &-5 &-4 \\ -3 & 0 & 1 \end{bmatrix}. It follows from the formula that the inverse matrix of $B$ is
$B^{-1}=\frac{1}{\det(B)}\Adj(B)=\frac{1}{5}\begin{bmatrix} -1 & 0 & 2 \\ -12 &5 &4 \\ 3 & 0 & -1 \end{bmatrix}.$

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### 1 Response

1. sehar says:

thanks i got my ans : )

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