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Let $$A = \begin{bmatrix} 2&3\\ -1&5\end{bmatrix}$$ and $$f(x) =x^2-4x+7$$ Show that $f(A)=O.$ Use this result to find $A^5$.

Solution:

$$f(A)=A^2-4A+7I\\= \begin{bmatrix} 2 & 3 \\ -1 & 2 \\ \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \\ \end{bmatrix}-4\begin{bmatrix} 2 & 3 \\ -1 & 2 \\ \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\=\begin{bmatrix} 4-3 & 6+6 \\ -2-2 & -3+4 \\ \end{bmatrix}-\begin{bmatrix} 8 & 12 \\ -4 & 8 \\ \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7\\ \end{bmatrix}\\=\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$$

How to find $A^5$ ?

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    What is the $(2,2)$ entry? $5$? $2$? Make up your mind. – 2017-02-21

4 Answers 4

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We have $$A^2 -4A +7I =O $$ $$\Rightarrow A^2 =4A-7I $$ $$\Rightarrow A^3=A^2 (A) =(4A-7I)A = 4A^2-7A = 4(4A-7I)-7A = 9A -28I $$ $$\Rightarrow A^5=A^3 (A^2) = A^3 (4A-7I) = (9A-28I)(4A-7I) = 36A^2 -175A +196I =36 (4A-7I)-175A+196I = -31A -56 I $$

Just repeatedly use the same formula while multiplying $A^2$ with $A$ continuously. Hope it helps.

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    Please explain how you got = 4(4A-7I)-7A. Rest i understood. – 2017-02-21
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    @bappy Observe that $A^2 =4A-7I$ – 2017-02-21
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    @ stud_iisc does he plugged the value of $$= A^2$$ – 2017-02-21
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Use polynomial division to write:

$$x^5=(x^2-4x+7)q(x)+r(x)$$ where $\deg r(x)<2$.

Then substitute $x=A$.

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    if it is a second degree polynomial then this should be the characteristic polynomial but it isn't? why ? just a question – 2017-02-21
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    If what is a second degree polynomial? – 2017-02-21
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Hint: Can you use said polynomial (Or a polynomial you can calculate from it) in order to find $A^5$ in terms of $A^4$ and $A^3$?

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Another, more advanced and abstract, way of looking at this is to observe that $\Bbb Z[A]\cong\Bbb Z[T]/(T^2-4T+7)$. To find what matrix goes to $T^5$, you just do Euclidean division on $T^5$, with the divisor $T^2-4T+7$. The remainder is your answer.