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Mathematics Question on Matrix

Let A=[2−1 11]A = \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix}. If the sum of the diagonal elements of A13A^{13} is 3n3^n, then nn is equal to _____.

Answer

The given matrix is:
A=[2−1 11].A = \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix}.
Compute successive powers of AA:
A2=[2−1 11][2−1 11]=[3−3 30].A^2 = \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\\ 3 & 0 \end{bmatrix}.
A3=A2⋅A=[3−3 30][2−1 11]=[3−6 6−3].A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -3 \\\ 3 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -6 \\\ 6 & -3 \end{bmatrix}.
A4=A3⋅A=[3−6 6−3][2−1 11]=[0−9 9−9].A^4 = A^3 \cdot A = \begin{bmatrix} 3 & -6 \\\ 6 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -9 \\\ 9 & -9 \end{bmatrix}.
A5=A4⋅A=[0−9 9−9][2−1 11]=[−9−9 9−18].A^5 = A^4 \cdot A = \begin{bmatrix} 0 & -9 \\\ 9 & -9 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -9 & -9 \\\ 9 & -18 \end{bmatrix}.
A6=A5⋅A=[−9−9 9−18][2−1 11]=[−270 0−27].A^6 = A^5 \cdot A = \begin{bmatrix} -9 & -9 \\\ 9 & -18 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -27 & 0 \\\ 0 & -27 \end{bmatrix}.
A7=A6⋅A=[−270 0−27][2−1 11]=[36−27 −2736].A^7 = A^6 \cdot A = \begin{bmatrix} -27 & 0 \\\ 0 & -27 \end{bmatrix} \begin{bmatrix} 2 & -1 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 36 & -27 \\\ -27 & 36 \end{bmatrix}.
Observe that the diagonal elements of A7A^7 are 3636 and 3636. Their sum is:
Sum of diagonal elements=36+36=72=32⋅35=37.\text{Sum of diagonal elements} = 36 + 36 = 72 = 3^2 \cdot 3^5 = 3^7.
Thus: n=7.n = 7.