If A = $\begin{bmatrix$}[r] 2 & 5 \ 1 & 3 \end{bmatrix}, B = \begin{bmatrix}[r] 4 & -2 \ -1 & 3 \end{bmatrix} and I is the identity matric of same order and A^t is transpose of matrix A, find A^t.B + BI.

$A = \begin{bmatrix$}[r] 2 & 5 \ 1 & 3 \end{bmatrix}, A^t = \begin{bmatrix}[r] 2 & 1 \ 5 & 3 \end{bmatrix}, I = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}.

$\Rightarrow A^t.B + BI = \begin{bmatrix$}[r] 2 & 1 \ 5 & 3 \end{bmatrix}\begin{bmatrix}[r] 4 & -2 \ -1 & 3 \end{bmatrix} + \begin{bmatrix}[r] 4 & -2 \ -1 & 3 \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix} + \begin{bmatrix}[r] 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \ -1 \times 1 + 3 \times 0 & -1 \times 0 + 3 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 8 - 1 & -4 + 3 \ 20 - 3 & -10 + 9 \end{bmatrix} + \begin{bmatrix}[r] 4 + 0 & 0 - 2 \ -1 + 0 & 0 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 7 & -1 \ 17 & -1 \end{bmatrix} + \begin{bmatrix}[r] 4 & -2 \ -1 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 7 + 4 & -1 + (-2) \ 17 + (-1) & -1 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 11 & -3 \ 16 & 2 \end{bmatrix}.

Hence, $A^t.B + BI = \begin{bmatrix$}[r] 11 & -3 \ 16 & 2 \end{bmatrix}.