Question

(i) Using Cayley Hamilton theorem, find 𝐴

8 − 𝐴

7 + 5𝐴

6 − 𝐴

5 + 𝐴

4 − 𝐴

3 + 6𝐴

2 +

𝐴 − 2𝐼 𝑖𝑓 [

1 2 −2

2 5 −4

3 7 −5

]

Answer 1

Given matrix is $A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix}$

Then according to Cayley Hamilton theorem,

$|A - \lambda I| = 0$

So we will have,

$A - \lambda I = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$|A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \end{vmatrix} = 0$

Then equation will be

$\lambda^3 - 5\lambda ^2+7\lambda -3=0$

According to Cayley Hamilton theorem,

Every matrix is the root of it's eigen matrix.

then, $A^3 - 5A^2+7A -3=0$ (1)

Given equation is $A^8 − 5A^7 + 7A^6 − 3A^5 + A^4 − 5A^3 + 8A^2 − 2A + I$

This equation can be written as,

$A^8 − 5A^7 + 7A^6 − 3A^5 + A^4 − 5A^3 + 8A^2 − 2A + I = (A^3 - 5A^2+7A -3)(A^5+A) + (A^2+A+I)$

From equation (1), above equation will be modified as,

$A^8 − 5A^7 + 7A^6 − 3A^5 + A^4 − 5A^3 + 8A^2 − 2A + I = A^2+A+I$

So putting value of $A^2,A,I$

we will get,

$A^8 − 5A^7 + 7A^6 − 3A^5 + A^4 − 5A^3 + 8A^2 − 2A + I$

$= \begin{bmatrix} 5 & 4 & 4 \\ 0 & 1 & 0 \\ 4 & 4 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}8&5&5\\ 0&3&0\\ 5&5&8\end{bmatrix}$