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Is the matrix A derogatory?

Justify your answer.

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

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Solution:

The above matrix is derogatory but while justification A doesn't turn out to be a zero matrix so we are taking another example to illustrate the method.

For example, $A=\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}$

Solution: $A=\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}$

$|A|=4$

C.E. equation is given by $|A-\lambda I|=0$

$\begin{vmatrix} 5-\lambda & -6 & -6 \\ -1 & 4-\lambda & 2 \\ 3 & -6 & -4-\lambda \end{vmatrix}=0$

$\lambda^{3} -(5) \lambda^{2} +(20-16-20-6+18+12)\lambda - 4=0$

$\lambda^{3} - 5 \lambda^{2}+8\lambda-4=0$

$\lambda=1,2,2$

Since the eigen values are repeated,

Therefore, the matrix is derogatory.

Justification:

Let us see whether the polynomial $(x-2)(x-1)=x^{2}-3x+2$ annhilates A.

By Calley Hamilton Theorem (CHT), x in terms of A.

$A^{2}-3A+2I=0$ ----------(1)

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

$A^{2} =A.A=\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}\begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}$

$A^{2}=\begin{bmatrix} 13 & -18 & -18 \\ -3 & 10 & 6 \\ 9 & -18 & -14 \end{bmatrix}$

$I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Put the above values in (1),

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

Therefore, Calley Hamilton Theorem is verified. Hence, the matrix is derogatory.

NOTE:

i) If the eigen values are repeated, then the given matrix is derogatory and verify using CHT, i.e., CHT=0.

ii) If eigen values are distinct, then the given matrix is non-derogatory and verify using CHT, i.e., CHT$\ne$0.

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