Find Inverse Matrices Using Adjoint Matrices

Problems and solutions in Linear Algebra

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

 
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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
\[B^{-1}=\frac{1}{\det(B)}\Adj(B)=\frac{1}{5}\begin{bmatrix}
-1 & 0 & 2 \\
-12 &5 &4 \\
3 & 0 & -1
\end{bmatrix}.\]


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