Answer to Question #151811 in Linear Algebra for adithya kumar

Question

given that a=[
1 0 3
2 1 -1
1 -1 1 ],compute the value of (a^6-5a^5+8a^4-2a^3-9a^2+31a-36 I ) using cayley-hamilton theorem

Answer 1

Characteristics equation is $|A-\lambda I|=0$.

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

$(1-\lambda) [(1-\lambda) ^2-1]+3[-2-(1-\lambda) ]=0\\ (1-\lambda) ^3-1+\lambda+3(-3+\lambda) =0\\ (1-\lambda) ^3-1+\lambda-9+3\lambda=0\\ (1-\lambda)^3+4\lambda-10=0\\ 1-3\lambda+3\lambda^2-\lambda^3+4\lambda-10=0\\ -\lambda^3+3\lambda^2+\lambda-9=0\\ \lambda^3-3\lambda^2-\lambda+9=0$

By Cayley-Hamilton Theorem, $A^3-3A^2-A+9=0.$

$\implies A^3-3A^2=A-9I.$

$A^6-5A^5+8A^4-2A^3-9A^2+31A-31I\\ =A^6-3A^5-2A^5+6A^4+2A^4-6A^3+4A^3-12A^2+3A^2+31A-36I\\ =A^3(A^3-3A^2)-2A^2(A^3-3A^2)+2A(A^3-3A^2)+4(A^3-3A^2)+3A^2+31A-36I\\ =(A^3-3A^2)(A^3-2A^2+2A+4I)+3A^2+31A-36I\\ =(A^3-3A^2)(A^3-3A^2+A^2-A+3A+9I-5I)+3A^2+31A-36I\\ =(A^3-3A^2)(A^3-3A^2-A+9I+A^2+3A-5I)+3A^2+31A-36I\\$

But, $(A^3-3A^2)=A-9I, A^3-3A^2-A+9I=0$

$=(A-9I)(0+A^2+3A-5)+3A^2+31A-36I\\ =A^3+3A^2-5A-9A^2-27A+45I+3A^2+31A-36I\\ =A^3-3A^2-A+9I\\ =0$