Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 6M

Year: Dec 2015

Characteristic eq. is

$A = \begin{bmatrix} 1 - \lambda & 2 \\ 2 & -1 - \lambda \end{bmatrix} = 0 \\ \therefore (1 - \lambda) (-1 - \lambda) - 4 = 0 \\ \therefore (\lambda^2 - 1) - 4 = 0 \\ \therefore \lambda^2 - 5 = 0$

According to cayley hamilton

$A^2 - 5I = 0 \\ \therefore A = \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} \\ A^2 = \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 2 & -1 \end{bmatrix} \\ = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

$\therefore A^2 - 5I \\ \therefore \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ = 0$

Hence verified

$A^4 = A^2 \cdot A^2 \\ A^4 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\ A^4 = \begin{bmatrix} 25 & 0 \\ 0 & 25 \end{bmatrix}$

Similarly

$A^{64} = \Bigg(\Big(\big((A^4)^2\big)^2\Big)^2\Bigg)^2 \\ \therefore A^{64} = \begin{bmatrix} 625^8 & 0 \\ 0 & 625^8 \end{bmatrix}$