Find the Inverse Matrix Using the Cayley-Hamilton Theorem
Problem 421
Find the inverse matrix of the matrix
\[A=\begin{bmatrix}
1 & 1 & 2 \\
9 &2 &0 \\
5 & 0 & 3
\end{bmatrix}\]
using the Cayley–Hamilton theorem.
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Solution.
To use the Cayley-Hamilton theorem, we first compute the characteristic polynomial $p(t)$ of the matrix $A$. We have
\begin{align*}
&p(t)=\det(A-tI)\\
&\begin{vmatrix}
1-t & 1 & 2 \\
9 &2-t &0 \\
5 & 0 & 3-t
\end{vmatrix}\\[6pt]
&=(-1)^{3+1}5\begin{vmatrix}
1 & 2\\
2-t& 0
\end{vmatrix}+(-1)^{3+2}\cdot 0 \begin{vmatrix}
1-t & 2\\
9& 0
\end{vmatrix}+(-1)^{3+3}(3-t)\begin{vmatrix}
1-t & 1\\
9& 2-t
\end{vmatrix} \\[6pt]
& \text{(by the 3rd row cofactor expansion)}\\
&=5(2t-4)+0+(3-t)\left(\, (1-t)(2-t)-9 \,\right)\\
&=-t^3+6t^2+8t-41.
\end{align*}
Then the Cayley-Hamilton theorem yields that $p(A)=O$, the zero matrix. That is, we have
\begin{align*}
O=p(A)=-A^3+6A^2+8A-41I.
\end{align*}
Thus, we have
\[41I=-A^3+6A^2+8A=A(-A^2+6A+8I),\]
or equivalently
\[I=A\left(\, \frac{1}{41}(-A^2+6A+8I) \,\right).\]
It follows that the inverse matrix is given by
\[A^{-1}=\frac{1}{41}(-A^2+6A+8I).\]
By a direct computation, we have
\[A^2=\begin{bmatrix}
20 & 3 & 8 \\
27 &13 &18 \\
20 & 5 & 19
\end{bmatrix}\]
and
\begin{align*}
-A^2+6A+8I&=-\begin{bmatrix}
20 & 3 & 8 \\
27 &13 &18 \\
20 & 5 & 19
\end{bmatrix}+6\begin{bmatrix}
1 & 1 & 2 \\
9 &2 &0 \\
5 & 0 & 3
\end{bmatrix}+8\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
-6 & 3 & 4 \\
27 &7 &-18 \\
10 & -5 & 7
\end{bmatrix}.
\end{align*}
Therefore the inverse matrix is
\[A^{-1}=\frac{1}{41}\begin{bmatrix}
-6 & 3 & 4 \\
27 &7 &-18 \\
10 & -5 & 7
\end{bmatrix}.\]
More Exercise
Test whether you understand how to find the inverse matrix using the Cayley-Hamilton theorem by the next problem.
\[A=\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}\] using the Cayley-Hamilton theorem.
The solution is given in the post “How to use the Cayley-Hamilton Theorem to Find the Inverse Matrix“.
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