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Let's say $f \in \mathbb{Q}[X]$ and $A \in \mathbb{Q}^{2 \times 2}$. Now let $f$ be $f = X^3+2X^2+3X+4$ and $A$ be $A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}$ how do we calculate the matrix $f(A)$?

I know from the solution, that the $(2,2)$ entry of $f(A)$ is $40$. However, I don't see how we' d calculate that. I've tried reversing the solution but I did not come far.

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    $\begingroup$ What is $A^2$? What is $A^3$? $\endgroup$ – lulu Jul 1 '19 at 20:52
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Since$$3A=\begin{bmatrix}3 & -3 \\ 6 & 9\end{bmatrix},$$since$$2A^2=\begin{bmatrix}-2 & -8 \\ 16 & 14\end{bmatrix},$$and since$$A^3=\begin{bmatrix}-9 & -11 \\ 22 & 13\end{bmatrix},$$we have$$A^3+2A^2+3A+4\operatorname{Id}=\begin{bmatrix}-4 & -22 \\ 44 & 40\end{bmatrix}.$$

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The simple approach is to follow the given calculation. Given $A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}$ we have $$A^2 = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}^2=\begin{pmatrix} -1 & -4 \\ 9 & 7 \end{pmatrix}$$ One more matrix multiply gets you $A^3$, then you multiply by the scalars and add to get $f(A)$. For higher powers you can often express $A=P^{-1}DP$ with $D$ diagonal to make the calculations easier because then $A^n=P^{-1}D^nP$ and diagonal matrices are easy to raise to a power.

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    $\begingroup$ If you are going to diagonalise $A$, chances are you are going to find an annihilating polynomial (like the characteristic one) along the way, any then you might as well reduce $f$ modulo that annihilating polynomial to reduce the degree. $\endgroup$ – Marc van Leeuwen Jul 3 '19 at 5:45
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If $f(x)=x^3+2x^2+3x+4,$ then $f(A)=A^3+2A^2+3A+4I.$ Can you compute this?

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