Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2015

Let $λ$ be Eigen value of matrix A $$\text{Characteristic equation is } |A-λI|=0$$

$$\therefore \begin{bmatrix} 7-λ & 4 & -1 \\ 4 & 7-λ & -1 \\ -4 & -4 & 4-λ \\ \end{bmatrix}=0$$

On solving, we get

$$λ^3-(7+7+4)λ^2+(24+24+33)λ-108=0$$

$$\therefore λ^3-18λ^2+81λ-108=0$$

$\therefore$ Eigen values $(λ)$ are $3,3,12$

Let $f(x)=(x-3)(x-12)=x^2+15x+36$

Now $A^2=A\times A$

$$=\begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \\ \end{bmatrix}\times \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \\ \end{bmatrix}$$

$$=\begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \\ \end{bmatrix}$$

$$\therefore A^2-15A+36I$$

$$=\begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \\ \end{bmatrix}-15\begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \\ \end{bmatrix}+36\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

$$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$=0$$

$\therefore f(x)=x^2-15x+36$ annihilates A

$\therefore f(x)$ is a minimal polynomial

Degree of $f(x)=2$ order of $A=3$

$\therefore$ Degree of $f(x)$ < order A

Hence, Matrix A is derogatory