Question

Given $M=\left[\begin{array}{cc}3& 2\\ -1& 4\end{array}\right],N=\left[\begin{array}{c}6\\ 1\end{array}\right],X=\left[\begin{array}{c}-4\\ 5\end{array}\right],Y=\left[\begin{array}{c}2\\ 2\end{array}\right]$, then find $MN+2X-4Y$

Answer 1

$MN+2X-4Y$

$=\left[\begin{array}{cc}3& 2\\ -1& 4\end{array}\right]\left[\begin{array}{c}6\\ 1\end{array}\right]+\left[\begin{array}{c}2\times -4\\ 2\times 5\end{array}\right]-\left[\begin{array}{c}4\times 2\\ 4\times 2\end{array}\right]$

$=\left[\begin{array}{c}3\times 6+2\times 1\\ -1\times 6+4\times 1\end{array}\right]+\left[\begin{array}{c}-8\\ 10\end{array}\right]-\left[\begin{array}{c}8\\ 8\end{array}\right]$

$=\left[\begin{array}{c}20\\ -2\end{array}\right]+\left[\begin{array}{c}-8\\ 10\end{array}\right]+\left[\begin{array}{c}-8\\ -8\end{array}\right]$

$=\left[\begin{array}{c}20-8-8\\ -2+10-8\end{array}\right]$

$=\left[\begin{array}{c}4\\ 0\end{array}\right]$

Hence, the answer is $\left[\begin{array}{c}4\\ 0\end{array}\right].$