Question
If $A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]$ and $I$ is Identity matrix of same order and $A^t$ is the transpose of matrix $A$ find $A^t \cdot B+B I$
✓
Answer
$A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right]$
$\therefore A^t=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$
$\begin{array}{l}A^t . B=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 \times 4+1 \times(-1) & 2 \times(-2)+1 \times 3 \\ 3 \times 4+3 \times(-1) & 5 \times(-2)+3 \times 3\end{array}\right] $
$ =\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]$
$B .I=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]$
$\therefore A^t \cdot B+B I=\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]+\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}11 & -3 \\ 16 & 2\end{array}\right]$
$\therefore A^t=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$
$\begin{array}{l}A^t . B=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 \times 4+1 \times(-1) & 2 \times(-2)+1 \times 3 \\ 3 \times 4+3 \times(-1) & 5 \times(-2)+3 \times 3\end{array}\right] $
$ =\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]$
$B .I=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]$
$\therefore A^t \cdot B+B I=\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]+\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}11 & -3 \\ 16 & 2\end{array}\right]$
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