Question: Chapter 2 .
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Check whether the following matrix is derogatory or non-derogatory.

$A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$

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modified 9 weeks ago  • written 9 weeks ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Solution:

$A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$

$|A|=32$

Characteristic Equation $|A- \lambda I|=0$

$\begin{vmatrix} 6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda \end{vmatrix}=0$

$\lambda^{3}-12\lambda^{2}+(18+9+18-4-4-1)\lambda -32=0$

$\lambda^{3}-12\lambda^{2}+36\lambda-32=0$

$\lambda=8, \ 2, \ 2$

Eigen values are 2, 2, 8.

$\therefore f(x)=(x-2)(x-8)=x^{2}-10x+16$

$\therefore f(A)=A^{2}-10A+16I$ ------------(1)

By Cayley-Hamilton Theorem,

$A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$ -----------(2)

$A^{2}=A.A=\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}=\begin{bmatrix} 44 & -20 & 20 \\ -20 & 14 & -10 \\ 20 & -10 & 14 \end{bmatrix}$ ------------(3)

$I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ -----------------(4)

Put (2), (3), (4) in equation (1),

$A^{2}-3A+2I=\begin{bmatrix} 44 & -20 & 20 \\ -20 & 14 & -10 \\ 20 & -10 & 14 \end{bmatrix}-10\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}+16\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Since Eigen values are rejected and Cayley-Hamilton Theorem is verified. Hence, the given matrix is derogatory.

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