Question #154528

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

-2 2


1
Expert's answer
2021-01-11T16:09:04-0500

Given matrix is A=[3122 ]A=\left [ \begin{matrix} 3 & 1 \\ -2 & 2 \end{matrix} \ \right ]

Find the characteristic polynomial equation for the given matrix AA.

A characteristic polynomial equation is given by

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

Using the given matrix, the characteristic equation is

3λ122λ =0\left | \begin{matrix} 3-\lambda & 1 \\ -2 & 2-\lambda \end{matrix} \ \right |=0

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

63λ2λ+λ2+2=0\Rightarrow 6-3\lambda -2\lambda +\lambda ^2+2=0

λ25λ+8=0\Rightarrow \lambda ^2-5\lambda +8=0

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

That is, we get A25A+8I=0\Rightarrow A ^2-5A +8I=0 ... (1)

Now let us take the given expression A5A4+A24IA^5-A^4+A^2-4I and express it as linear polynomial using the equation (1)

Now A5A4+A24I=A3(A25A+8I)+4A48A3+A2A^5-A^4+A^2-4I=A^3(A^2-5A+8I)+4A^4-8A^3+A^2

=A3(0)+4A48A3+A2=A^3(0)+4A^4-8A^3+A^2 ( using (1) )

=4A48A3+A2=4A^4-8A^3+A^2

=4A2(A25A+8I)+12A331A2=4A^2(A^2-5A+8I)+12A^3-31A^2

=4A2(0)+12A331A2=4A^2(0)+12A^3-31A^2

=12A331A2=12A^3-31A^2

=12A(A25A+8I)+29A2=12A(A^2-5A+8I)+29A^2

=12A(0)+29A2=12A(0)+29A^2

=29A2=29A^2

=29(A25A+8I)+145A232I=29(A^2-5A+8I)+145A-232I

=29(0)+145A232I=29(0)+145A-232I

=145A232I=145A-232I

Therefore, A5A4+A24I=145A232IA^5-A^4+A^2-4I=145A-232I


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