Question

Let A = $\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]$, B = $\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right]$ and C = $\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right]$. Find each of the following:
(i) 2A − 3B
(ii) B − 4C
(iii) 3A − C
(iv) 3A − 2B + 3C

Answer 1

$$\begin{gathered} \left(i\right)2A-3B \\ \Rightarrow 2A-3B=2\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]-3\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right] \\ \Rightarrow 2A-3B=\left[\begin{array}{cc}4& 8\\ 6& 4\end{array}\right]-\left[\begin{array}{cc}3& 9\\ -6& 15\end{array}\right] \\ \Rightarrow 2A-3B=\left[\begin{array}{cc}4-3& 8-9\\ 6+6& 4-15\end{array}\right] \\ \Rightarrow 2A-3B=\left[\begin{array}{cc}1& -1\\ 12& -11\end{array}\right] \\ \left(ii\right)B-4C \\ \Rightarrow B-4C=\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right]-4\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right] \\ \Rightarrow B-4C=\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right]-\left[\begin{array}{cc}-8& 20\\ 12& 16\end{array}\right] \\ \Rightarrow B-4C=\left[\begin{array}{cc}1+8& 3-20\\ -2-12& 5-16\end{array}\right] \\ \Rightarrow B-4C=\left[\begin{array}{cc}9& -17\\ -14& -11\end{array}\right] \\ \left(iii\right)3A-C \\ \Rightarrow 3A-C=3\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]-\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right] \\ \Rightarrow 3A-C=\left[\begin{array}{cc}6& 12\\ 9& 6\end{array}\right]-\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right] \\ \Rightarrow 3A-C=\left[\begin{array}{cc}6+2& 12-5\\ 9-3& 6-4\end{array}\right] \\ \Rightarrow 3A-C=\left[\begin{array}{cc}8& 7\\ 6& 2\end{array}\right] \\ \left(iv\right)3A-2B+3C \\ \Rightarrow 3A-2B+3C=3\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right]-2\left[\begin{array}{cc}1& 3\\ -2& 5\end{array}\right]+3\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right] \\ \Rightarrow 3A-2B+3C=\left[\begin{array}{cc}6& 12\\ 9& 6\end{array}\right]-\left[\begin{array}{cc}2& 6\\ -4& 10\end{array}\right]+\left[\begin{array}{cc}-6& 15\\ 9& 12\end{array}\right] \\ \Rightarrow 3A-2B+3C=\left[\begin{array}{cc}6-2-6& 12-6+15\\ 9+4+9& 6-10+12\end{array}\right] \\ \Rightarrow 3A-2B+3C=\left[\begin{array}{cc}-2& 21\\ 22& 8\end{array}\right] \end{gathered}$$