Show that the matrix:

$A=\begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$

Also find the diagonal and transforming matrix.

Solution:

$A=\begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$

$|A|=3$

To find eigen values:

$|A- \lambda I|=0$

$A=\begin{vmatrix} -9- \lambda & 4 & 4 \\ -8 & 3- \lambda & 4 \\ -16 & 8 & 7- \lambda \end{vmatrix}=0$

$\lambda^{3}-(-9+3+7)\lambda^{2}+(-27+21-63+32+64-32)\lambda-3=0$

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

$\lambda=-1, -1,3$

To find the eigen vector,

$[A-\lambda I]X=0$

$\begin{bmatrix} -9- \lambda & 4 & 4 \\ -8 & 3- \lambda & 4 \\ -16 & 8 & 7- \lambda \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ ---------(1)

Putting $\lambda=-1$ in (1),

$\begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Therefore, $R_{2}-R_{1} $ and $R_{3}-2R_{1}$,

$\begin{bmatrix} -8 & 4 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Therefore, $\cfrac{R_{1}}{4}$,

$\begin{bmatrix} -2 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$-2x_{1}+x_{2}+x_{3}=0$ ---------(2)

$\therefore$ Rank of the matrix = 7

Put $x_{2}=2s$ and $x_{3}=2t$ in equation (2),

$-2x_{1}+2s+2t=0$

$2x_{1}=2s+2t$

$x_{1}=s+t$

Now from equation (2),

$x_{2} = 2x_{1}-x_{3}$

$x_{2}=2(s+t)-2t$

$x_{2}=2s+2t-2t$

$x_{2}=2s$

Similarly, from equation (2),

$x_{3}=2x_{1}-x_{2}$

$x_{3}=2(s+t)-2s$

$x_{3}=2s+2t-2s$

$x_{3}=2t$

$X_{1}=\begin{bmatrix} s+t \\ 2s+0 \\ 0+2t \end{bmatrix}=\begin{bmatrix} s \\ 2s \\ 0 \end{bmatrix}+\begin{bmatrix} t \\ 0 \\ 2t \end{bmatrix}$

$X_{1}=s \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}+t\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$

$X_{1}=[1, 2, 0]$ ---------(3)

$X_{1}=[1, 0, 2]$ ---------(4)

Put $\lambda=3$ in (1),

$\begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$-12x_{1}+4x_{2}+4x_{3}=0$

$-8x_{1}+0x_{2}+4x_{3}=0$

$-16x_{1}+8x_{2}+4x_{3}=0$

$\cfrac{x_{1}}{\begin{vmatrix} 4 & 4 \\ 0 & 4 \end{vmatrix}}=-\cfrac{x_{2}}{\begin{vmatrix} -12 & 4 \\ -8 & 4 \end{vmatrix}}=\cfrac{x_{3}}{\begin{vmatrix} -12 & 4 \\ -8 & 0 \end{vmatrix}}$

$\cfrac{x_{1}}{16-0}=\cfrac{x_{2}}{-48+32}=\cfrac{x_{3}}{0+32}$

$\cfrac{x_{1}}{16}=\cfrac{x_{2}}{-16}=\cfrac{x_{3}}{32}$

$X_{3}=\begin{bmatrix} 16 \\ 16 \\ 32 \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$

$X_{3}=[1, 1, 2]$ ---------(5)

$\therefore M=[X_{1} \ X_{2} \ X_{3} ]$,

$M=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & 2 & 2 \end{bmatrix}$

Since $M^{-1}AM=D$

$A=\begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}$ will be diagonalised to the diagonal matrix

$D=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

by the transforming matrix $M=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & 2 & 2 \end{bmatrix}$