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

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

 
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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
\end{vmatrix}=-1,\quad C_{12}=-\begin{vmatrix}
0 & 2\\
0& 1
\end{vmatrix}=0, \quad C_{13}=\begin{vmatrix}
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
\end{vmatrix}=1, \quad C_{12}=-\begin{vmatrix}
0 & 4\\
3& 1
\end{vmatrix}=12. \quad C_{13}=\begin{vmatrix}
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
\[B^{-1}=\frac{1}{\det(B)}\Adj(B)=\frac{1}{5}\begin{bmatrix}
-1 & 0 & 2 \\
-12 &5 &4 \\
3 & 0 & -1
\end{bmatrix}.\] Click here if solved 142

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