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 *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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