Answer to Question #154528 in Linear Algebra for Rocky

Question

Express A^5-A^4+A^2-4I as linear polynomial in A where A=metrix 3 1

-2 2

Answer 1

Given matrix is $A=\left [ \begin{matrix} 3 & 1 \\ -2 & 2 \end{matrix} \ \right ]$

Find the characteristic polynomial equation for the given matrix $A$.

A characteristic polynomial equation is given by

$\left | A-\lambda I \right |=0$, where $\lambda$ is any scalar.

Using the given matrix, the characteristic equation is

$\left | \begin{matrix} 3-\lambda & 1 \\ -2 & 2-\lambda \end{matrix} \ \right |=0$

$\Rightarrow (3-\lambda )(2-\lambda )+2=0$

$\Rightarrow 6-3\lambda -2\lambda +\lambda ^2+2=0$

$\Rightarrow \lambda ^2-5\lambda +8=0$

Caley Hamilitons theorem states that every square matrix satisfies its characteristic equation.

That is, we get $\Rightarrow A ^2-5A +8I=0$ ... (1)

Now let us take the given expression $A^5-A^4+A^2-4I$ and express it as linear polynomial using the equation (1)

Now $A^5-A^4+A^2-4I=A^3(A^2-5A+8I)+4A^4-8A^3+A^2$

$=A^3(0)+4A^4-8A^3+A^2$ ( using (1) )

$=4A^4-8A^3+A^2$

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

$=4A^2(0)+12A^3-31A^2$

$=12A^3-31A^2$

$=12A(A^2-5A+8I)+29A^2$

$=12A(0)+29A^2$

$=29A^2$

$=29(A^2-5A+8I)+145A-232I$

$=29(0)+145A-232I$

$=145A-232I$

Therefore, $A^5-A^4+A^2-4I=145A-232I$