Given matrix is A = [ 3 1 − 2 2 ] A=\left [ \begin{matrix}
3 & 1 \\
-2 & 2
\end{matrix}
\ \right ] A = [ 3 − 2 1 2 ]
Find the characteristic polynomial equation for the given matrix A A A .
A characteristic polynomial equation is given by
∣ A − λ I ∣ = 0 \left | A-\lambda I \right |=0 ∣ A − λ I ∣ = 0 , where λ \lambda λ is any scalar.
Using the given matrix, the characteristic equation is
∣ 3 − λ 1 − 2 2 − λ ∣ = 0 \left |
\begin{matrix}
3-\lambda & 1 \\
-2 & 2-\lambda
\end{matrix}
\ \right |=0 ∣ ∣ 3 − λ − 2 1 2 − λ ∣ ∣ = 0
⇒ ( 3 − λ ) ( 2 − λ ) + 2 = 0 \Rightarrow (3-\lambda )(2-\lambda )+2=0 ⇒ ( 3 − λ ) ( 2 − λ ) + 2 = 0
⇒ 6 − 3 λ − 2 λ + λ 2 + 2 = 0 \Rightarrow 6-3\lambda -2\lambda +\lambda ^2+2=0 ⇒ 6 − 3 λ − 2 λ + λ 2 + 2 = 0
⇒ λ 2 − 5 λ + 8 = 0 \Rightarrow \lambda ^2-5\lambda +8=0 ⇒ λ 2 − 5 λ + 8 = 0
Caley Hamilitons theorem states that every square matrix satisfies its characteristic equation.
That is, we get ⇒ A 2 − 5 A + 8 I = 0 \Rightarrow A ^2-5A +8I=0 ⇒ A 2 − 5 A + 8 I = 0 ... (1)
Now let us take the given expression A 5 − A 4 + A 2 − 4 I A^5-A^4+A^2-4I A 5 − A 4 + A 2 − 4 I and express it as linear polynomial using the equation (1)
Now A 5 − A 4 + A 2 − 4 I = A 3 ( A 2 − 5 A + 8 I ) + 4 A 4 − 8 A 3 + A 2 A^5-A^4+A^2-4I=A^3(A^2-5A+8I)+4A^4-8A^3+A^2 A 5 − A 4 + A 2 − 4 I = A 3 ( A 2 − 5 A + 8 I ) + 4 A 4 − 8 A 3 + A 2
= A 3 ( 0 ) + 4 A 4 − 8 A 3 + A 2 =A^3(0)+4A^4-8A^3+A^2 = A 3 ( 0 ) + 4 A 4 − 8 A 3 + A 2 ( using (1) )
= 4 A 4 − 8 A 3 + A 2 =4A^4-8A^3+A^2 = 4 A 4 − 8 A 3 + A 2
= 4 A 2 ( A 2 − 5 A + 8 I ) + 12 A 3 − 31 A 2 =4A^2(A^2-5A+8I)+12A^3-31A^2 = 4 A 2 ( A 2 − 5 A + 8 I ) + 12 A 3 − 31 A 2
= 4 A 2 ( 0 ) + 12 A 3 − 31 A 2 =4A^2(0)+12A^3-31A^2 = 4 A 2 ( 0 ) + 12 A 3 − 31 A 2
= 12 A 3 − 31 A 2 =12A^3-31A^2 = 12 A 3 − 31 A 2
= 12 A ( A 2 − 5 A + 8 I ) + 29 A 2 =12A(A^2-5A+8I)+29A^2 = 12 A ( A 2 − 5 A + 8 I ) + 29 A 2
= 12 A ( 0 ) + 29 A 2 =12A(0)+29A^2 = 12 A ( 0 ) + 29 A 2
= 29 A 2 =29A^2 = 29 A 2
= 29 ( A 2 − 5 A + 8 I ) + 145 A − 232 I =29(A^2-5A+8I)+145A-232I = 29 ( A 2 − 5 A + 8 I ) + 145 A − 232 I
= 29 ( 0 ) + 145 A − 232 I =29(0)+145A-232I = 29 ( 0 ) + 145 A − 232 I
= 145 A − 232 I =145A-232I = 145 A − 232 I
Therefore, A 5 − A 4 + A 2 − 4 I = 145 A − 232 I A^5-A^4+A^2-4I=145A-232I A 5 − A 4 + A 2 − 4 I = 145 A − 232 I
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