Question

If $A=\left[\begin{array}{cc}3& 5\\ 7& -9\end{array}\right]$ and $B=\left[\begin{array}{cc}6& -4\\ 2& 3\end{array}\right]$, find $(4A-3B)$.

Answer 1

$$\begin{array}{c}A=\left[\begin{array}{cc}3& 5\\ 7& -9\end{array}\right]andB=\left[\begin{array}{cc}6& -4\\ 2& 3\end{array}\right]\\ \left(4A-3B\right)=4A+\left(-3B\right)\\ Now,\\ 4A=\left[\begin{array}{cc}4\times 3& 4\times 5\\ 4\times 7& 4\times \left(-9\right)\end{array}\right]=\\ =\left[\begin{array}{cc}12& 20\\ 28& -36\end{array}\right]\\ And,\\ -3B=\left(-3\right)\times B=\left[\begin{array}{cc}\left(-3\right)\times 6& \left(-3\right)\times \left(-4\right)\\ \left(-3\right)\times 2& \left(-3\right)\times 3\end{array}\right]\\ =\left[\begin{array}{cc}-18& 12\\ -6& -9\end{array}\right]\\ \therefore \left(4A-3B\right)=4A+\left(-3B\right)\\ =\left[\begin{array}{cc}12& 20\\ 28& -36\end{array}\right]+\left[\begin{array}{cc}-18& 12\\ -6& -9\end{array}\right]\\ =\left[\begin{array}{cc}12+\left(-18\right)& 20+12\\ 28+\left(-6\right)& -36+\left(-9\right)\end{array}\right]\\ =\left[\begin{array}{cc}-6& 32\\ 22& -45\end{array}\right]\\ Hence,\\ \left(4A-3B\right)=\left[\begin{array}{cc}-6& 32\\ 22& -45\end{array}\right]\end{array}$$