Answer to Question #153071 in Linear Algebra for Kishore Kris

Question

2. ſi 0 Verify Cayley-Hamilton theorem, Find A ^ 4 when A= 2 1 1 1 1 -1 3 -1 1

Answer 1

$A=\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}$

Characteristic polynomial of A,

$det(A-\lambda I)=0$

$\Rightarrow det\left(\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix} - \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\right)=0$

$\Rightarrow det \left(\begin{bmatrix}2-\lambda&1&1\\1&1-\lambda&-1\\3&-1&1-\lambda\end{bmatrix}\right)=0$

$\Rightarrow (2-\lambda)det\left(\begin{bmatrix}1-\lambda&-1\\-1&1-\lambda\end{bmatrix}\right)$

$-det\left(\begin{bmatrix}1&-1\\3&1-\lambda\end{bmatrix}\right)$

$+det\left(\begin{bmatrix}1&1-\lambda\\3&-1\end{bmatrix}\right)=0$

$\Rightarrow \left(2-\lambda\right)\left(\lambda^2-2\lambda\right)-\left(-\lambda+4\right)+ \left(3\lambda-4\right)=0$

$\Rightarrow \lambda^3-4\lambda^2+8=0$

$\therefore$ Characteristic polynomial of A :- $\lambda^3-4\lambda^2+8=0$

Now,

$A^2= A\times A =\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}\times \begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}$

$=\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}$

and,

$A^3=A^2\times A=\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}\times \begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}$

$=\begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}$

Now, by Cayley-Hamilton theorem, putting $\lambda=A$ ,

$A^3-4A^2+8I=0_{3\times3}$

$\Rightarrow \begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}-4\begin{bmatrix}8&2&2\\0&3&-1\\8&1&5\end{bmatrix}$

$+8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=0_{3\times3}$

$\Rightarrow\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0_{3\times3}$

$\therefore$ Cayley-Hamilton Theorem Verified.

Now, by Cayley-Hamilton theorem,

$A^3-4A^2+8I=0_{3\times3}$

$\Rightarrow A(A^3-4A^2+8I)=A\times0_{3\times3}$

[ Multiplying both side by matrix $A$ ]

$\Rightarrow A^4-4A^3+8A=0_{3\times3}$

$\Rightarrow A^4 = 4A^3-8A$

$=4\begin{bmatrix}24&8&8\\0&4&-4\\32&4&12\end{bmatrix}-8\begin{bmatrix}2&1&1\\1&1&-1\\3&-1&1\end{bmatrix}$ $=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}$

$\therefore A^4=\begin{bmatrix}80&24&24\\-8&8&-8\\104&24&40\end{bmatrix}$