Question

Let P = [ -1 4 5] . Determine P^-1 using

[ 0 2 -3]

[ 0 0 8]

Cayley- Hamilton theorem. Further use P^-1 to express (x1, x2, x3) in terms of (-1,0,0), (4,2,0), ( 5,-3,8)

Answer 1

$P=\begin{bmatrix} -1&4&5 \\ 0&2&-3 \\ 0&0&8 \end{bmatrix}$

$p(\lambda)=\det (P-\lambda I)=\begin{vmatrix} -1-\lambda&4&5 \\ 0&2-\lambda &-3 \\ 0&0&8-\lambda \end{vmatrix} =(-1-\lambda )(2-\lambda )(8-\lambda)$

$p(\lambda )=-x^3+9x^2-6x-16$

$-P^3+9P^2-6P-16I=0$

$P(-P^2+9P-6I)=16I$

$16P^{-1}=-P^2+9P-6I=-\begin{bmatrix} 1&4&23 \\ 0&4&-30 \\ 0&0&64 \end{bmatrix}+9\begin{bmatrix} -1&4&5 \\ 0&2&-3 \\ 0&0&8 \end{bmatrix}-\begin{bmatrix} 6&0&0 \\ 0&6&0 \\ 0&0&6 \end{bmatrix}$

$16P^{-1}=\begin{bmatrix} -16&32&22 \\ 0&8&3 \\ 0&0&2 \end{bmatrix}$

$P^{-1}=\frac{1}{16}\begin{bmatrix} -16&32&22 \\ 0&8&3 \\ 0&0&2 \end{bmatrix}$

Let $\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}=a\cdot \begin{bmatrix}-1\\0\\0 \end{bmatrix}+ b\cdot \begin{bmatrix}4\\2\\0 \end{bmatrix}+ c\cdot \begin{bmatrix}5\\-3\\8 \end{bmatrix}=P\begin{bmatrix}a\\b\\c\end{bmatrix}$ .

Then $\begin{bmatrix} a\\b\\c \end{bmatrix}=P^{-1}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix} -x_1+2x_2+\tfrac{11}{8} x_3 \\ \tfrac{1}{2}x_2+\tfrac{3}{16}x_3 \\ \tfrac{1}{8}x_3 \end{bmatrix}$ .

So, $\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}=(-x_1+2x_2+\tfrac{11}{8}x_3)\cdot \begin{bmatrix}-1\\0\\0 \end{bmatrix}+ (\tfrac{1}{2}x_2+\tfrac{3}{16}x_3)\cdot \begin{bmatrix}4\\2\\0 \end{bmatrix}+ \tfrac{1}{8}x_3\cdot \begin{bmatrix}5\\-3\\8 \end{bmatrix}$ .

Answers:

$P^{-1}=\frac{1}{16}\begin{bmatrix} -16&32&22 \\ 0&8&3 \\ 0&0&2 \end{bmatrix}$

$\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}=(-x_1+2x_2+\tfrac{11}{8}x_3)\cdot \begin{bmatrix}-1\\0\\0 \end{bmatrix}+ (\tfrac{1}{2}x_2+\tfrac{3}{16}x_3)\cdot \begin{bmatrix}4\\2\\0 \end{bmatrix}+ \tfrac{1}{8}x_3\cdot \begin{bmatrix}5\\-3\\8 \end{bmatrix}$