written 5.0 years ago by | modified 2.1 years ago by |
Justify your answer.
where $A=\begin{bmatrix} -2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$
written 5.0 years ago by | modified 2.1 years ago by |
Justify your answer.
where $A=\begin{bmatrix} -2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$
written 5.0 years ago by |
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.
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.