Given matrix is A=[3−212 ]
Find the characteristic polynomial equation for the given matrix A.
A characteristic polynomial equation is given by
∣A−λI∣=0, where λ is any scalar.
Using the given matrix, the characteristic equation is
∣∣3−λ−212−λ ∣∣=0
⇒(3−λ)(2−λ)+2=0
⇒6−3λ−2λ+λ2+2=0
⇒λ2−5λ+8=0
Caley Hamilitons theorem states that every square matrix satisfies its characteristic equation.
That is, we get ⇒A2−5A+8I=0 ... (1)
Now let us take the given expression A5−A4+A2−4I and express it as linear polynomial using the equation (1)
Now A5−A4+A2−4I=A3(A2−5A+8I)+4A4−8A3+A2
=A3(0)+4A4−8A3+A2 ( using (1) )
=4A4−8A3+A2
=4A2(A2−5A+8I)+12A3−31A2
=4A2(0)+12A3−31A2
=12A3−31A2
=12A(A2−5A+8I)+29A2
=12A(0)+29A2
=29A2
=29(A2−5A+8I)+145A−232I
=29(0)+145A−232I
=145A−232I
Therefore, A5−A4+A2−4I=145A−232I
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