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Show that the matrix $$ A= \begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \\ \end{bmatrix}$$ is derogatory.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2014

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

$$\therefore \begin{bmatrix} 5-λ & -6 & -6 \\ -1 & 4-λ & 2 \\ 3 & -6 & -4-λ \\ \end{bmatrix}=0$$

on solving we get

$$λ^3-(5+4-4)λ^2+(-4-2+14)λ-4=0$$

$\therefore λ^3-5λ^2+8λ-4=0$

$\therefore $ Eigen value$(λ)$ are $1,2,2$

Let $f(x)=(x-1)(x-2)=x^2-3x+2$

Now $A^2=A\times A$

$$=\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \\ \end{bmatrix}=0\times \begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \\ \end{bmatrix}=\begin{bmatrix} 13 & -18 & -18 \\ -3 & 10 & 6 \\ 9 & -18 & -14 \\ \end{bmatrix}$$

$$\therefore A^2-3A+2I$$

$$=\begin{bmatrix} 13 & -18 & -18 \\ -3 & 10 & 6 \\ 9 & -18 & -14 \\ \end{bmatrix}-3\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \\ \end{bmatrix}+2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

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

$$=0$$

$\therefore f(x)=x^2-3x+2$ annihilates A

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

Degree of f(x) < order of A

$\therefore $ Matrix A is derogatory

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