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Let $A= \begin{pmatrix} 1 & 0 & 2 \\ 1 & -2 & 0 \\ 0 & 0 & - 3 \\ \end{pmatrix} $ and $I$ be the identity matrix. If $6A^-=aA^2+bA+cI$ where $a,b,c$ are real numbers, then obtain the value of $(a,b,c).$

Options are: a) (1,2,1), b) (1,-1,2), c) (4,1,1), d) (1,4,1)

Calculate this with proper explanation and minimal working please. I am new to this topic. Thanks in advance.

S.S
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1 Answers1

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If $p(t)=a_3t^3+a_2t^2+a_1t+a_0$ is the char. polynomial of $A$, then , by Cayley -Hamilton:

$a_3A^3+a_2A^2+a_1A+a_0I=0$.

Compute $a_3,a_2,a_1$ and $a_0$. You will see thatb $a_0 \ne 0$.

This gives:

$a_0A^{-1}=a_3A^2+a_2A+a_1I$

Your turn !

Fred
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  • I solved the characteristic equation and after substituting the matrix $A$ in it (using Caley Hamilton theorem)I got $$A^3+4A^2+A=6I$$; then I operated by $A^{-1}$ on both sides to obtain $$(a,b,c)=(1,4,1)$$. That solved my question. Thanks a lot sir! – S.S Nov 13 '17 at 18:41