Let A=begin{bmatrix} 2 & -1 1 & 1 end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix}$.Calculate the characteristic equation of $A$: $|A - \lambda I| = 0$.$\begin{vmatrix} 2-\

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$7$

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(A) Given $A = \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix}$.Calculate the characteristic equation of $A$: $|A - \lambda I| = 0$.$\begin{vmatrix} 2-\lambda & -1 \ 1 & 1-\lambda \end{vmatrix} = (2-\lambda)(1-\lambda) + 1 = \lambda^2 - 3\lambda + 3 = 0$.By Cayley-Hamilton theorem,$A^2 - 3A + 3I = 0$,so $A^2 = 3A - 3I$.We observe $A^6 = -27I = -3^3 I$.Then $A^{12} = (A^6)^2 = (-3^3 I)^2 = 3^6 I$.$A^{13} = A^{12} \cdot A = 3^6 A = 3^6 \begin{bmatrix} 2 & -1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 3^6 & -3^6 \ 3^6 & 3^6 \end{bmatrix}$.The sum of the diagonal elements (trace) of $A^{13}$ is $2 \cdot 3^6 + 3^6 = 3 \cdot 3^6 = 3^7$.Given the sum is $3^n$,we have $3^n = 3^7$,which implies $n = 7$.

Let $A=\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$. If the sum of the diagonal elements of $A^{13}$ is $3^{n}$,then $n$ is equal to ..........

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