Find the Inverse Matrix Using the Cayley-Hamilton Theorem

Cayley-Hamilton Theorem Problems and Solutions

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.

 
LoadingAdd to solve later
Sponsored Links

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.

Problem. Find the inverse matrix of the $3\times 3$ matrix
\[A=\begin{bmatrix}
7 & 2 & -2 \\
-6 &-1 &2 \\
6 & 2 & -1
\end{bmatrix}\] using the Cayley-Hamilton theorem.

The solution is given in the post “How to use the Cayley-Hamilton Theorem to Find the Inverse Matrix“.

More Problems about the Cayley-Hamilton Theorem

Problems about the Cayley-Hamilton theorem and their solutions are collected on the page:

The Cayley-Hamilton Theorem


LoadingAdd to solve later

Sponsored Links

More from my site

  • How to Use the Cayley-Hamilton Theorem to Find the Inverse MatrixHow to Use the Cayley-Hamilton Theorem to Find the Inverse Matrix Find the inverse matrix of the $3\times 3$ matrix \[A=\begin{bmatrix} 7 & 2 & -2 \\ -6 &-1 &2 \\ 6 & 2 & -1 \end{bmatrix}\] using the Cayley-Hamilton theorem.   Solution. To apply the Cayley-Hamilton theorem, we first determine the characteristic […]
  • Find All the Eigenvalues of Power of Matrix and Inverse MatrixFind All the Eigenvalues of Power of Matrix and Inverse Matrix Let \[A=\begin{bmatrix} 3 & -12 & 4 \\ -1 &0 &-2 \\ -1 & 5 & -1 \end{bmatrix}.\] Then find all eigenvalues of $A^5$. If $A$ is invertible, then find all the eigenvalues of $A^{-1}$.   Proof. We first determine all the eigenvalues of the matrix […]
  • Eigenvalues and their Algebraic Multiplicities of a Matrix with a VariableEigenvalues and their Algebraic Multiplicities of a Matrix with a Variable Determine all eigenvalues and their algebraic multiplicities of the matrix \[A=\begin{bmatrix} 1 & a & 1 \\ a &1 &a \\ 1 & a & 1 \end{bmatrix},\] where $a$ is a real number.   Proof. To find eigenvalues we first compute the characteristic polynomial of the […]
  • Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam)Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam) Let \[\begin{bmatrix} 0 & 0 & 1 \\ 1 &0 &0 \\ 0 & 1 & 0 \end{bmatrix}.\] (a) Find the characteristic polynomial and all the eigenvalues (real and complex) of $A$. Is $A$ diagonalizable over the complex numbers? (b) Calculate $A^{2009}$. (Princeton University, […]
  • Rotation Matrix in Space and its Determinant and EigenvaluesRotation Matrix in Space and its Determinant and Eigenvalues For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by \[A=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta &\cos\theta &0 \\ 0 & 0 & 1 \end{bmatrix}.\] (a) Find the determinant of the matrix $A$. (b) Show that $A$ is an […]
  • If 2 by 2 Matrices Satisfy $A=AB-BA$, then $A^2$ is Zero MatrixIf 2 by 2 Matrices Satisfy $A=AB-BA$, then $A^2$ is Zero Matrix Let $A, B$ be complex $2\times 2$ matrices satisfying the relation \[A=AB-BA.\] Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix.   Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […]
  • Maximize the Dimension of the Null Space of $A-aI$Maximize the Dimension of the Null Space of $A-aI$ Let \[ A=\begin{bmatrix} 5 & 2 & -1 \\ 2 &2 &2 \\ -1 & 2 & 5 \end{bmatrix}.\] Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix. Your score of this problem is equal to that […]
  • Find Inverse Matrices Using Adjoint MatricesFind Inverse Matrices Using Adjoint Matrices 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 […]

You may also like...

1 Response

  1. 07/07/2017

    […] The solution is given in the post “Find the Inverse Matrix Using the Cayley-Hamilton Theorem“. […]

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue

(a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$...

Close