Question

Compute the products AB and BA whichever exists in each of the following cases:
(i) $A=\left[\begin{array}{cc}1& -2\\ 2& 3\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right]$

(ii) $A=\left[\begin{array}{cc}3& 2\\ -1& 0\\ -1& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}4& 5& 6\\ 0& 1& 2\end{array}\right]$
(iii) A = [1 −1 2 3] and $B=\left[\begin{array}{c}0\\ 1\\ 3\\ 2\end{array}\right]$

(iv) [a, b]$\left[\begin{array}{c}c\\ d\end{array}\right]$ + [a, b, c, d]$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ AB=\left[\begin{array}{cc}1& -2\\ 2& 3\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}1-4& 2-6& 3-2\\ 2+6& 4+9& 6+3\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}-3& -4& 1\\ 8& 13& 9\end{array}\right] \end{gathered}$$

Since the number of columns in B is greater then the number of rows in A, BA does not exists.

$$\begin{gathered} \left(\mathrm{ii}\right) \\ AB=\left[\begin{array}{cc}3& 2\\ -1& 0\\ -1& 1\end{array}\right]\left[\begin{array}{ccc}4& 5& 6\\ 0& 1& 2\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}12+0& 15+2& 18+4\\ -4+0& -5+0& -6+0\\ -4+0& -5+1& -6+2\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}12& 17& 22\\ -4& -5& -6\\ -4& -4& -4\end{array}\right] \\ \mathrm{Also}, \\ BA=\left[\begin{array}{ccc}4& 5& 6\\ 0& 1& 2\end{array}\right]\left[\begin{array}{cc}3& 2\\ -1& 0\\ -1& 1\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{cc}12-5-6& 8+0+6\\ 0-1-2& 0+0+2\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{cc}1& 14\\ -3& 2\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ AB=\left[\begin{array}{cccc}1& -1& 2& 3\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 3\\ 2\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{c}0+\left(-1\right)+6+6\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{c}11\end{array}\right] \\ \mathrm{Also}, \\ BA=\left[\begin{array}{c}0\\ 1\\ 3\\ 2\end{array}\right]\left[\begin{array}{cccc}1& -1& 2& 3\end{array}\right] \\ \Rightarrow BA=\left[\begin{array}{cccc}0& 0& 0& 0\\ 1& -1& 2& 3\\ 3& -3& 6& 9\\ 2& -2& 4& 6\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \\ \left[\begin{array}{cc}a& b\end{array}\right]\left[\begin{array}{c}c\\ d\end{array}\right]+\left[\begin{array}{cccc}a& b& c& d\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}ac+bd\end{array}\right]+\left[\begin{array}{c}{a}^2+b^2+c^2+d^2\end{array}\right] \\ \left[\begin{array}{c}{a}^2+b^2+c^2+d^2+ac+bd\end{array}\right] \end{gathered}$$