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If matrix A=[2312]A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, then A2+IA^2+I is equal to: (Where I is an identity matrix of order 2)

Solution

✅ Correct Option: 2

A2=[71247]A^2 = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}, so A2+I=[81248]=4[2312]=4AA^2+I = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} = 4\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = 4A.

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