Question

Let $A=\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right]\mathrm{and}B=\left[\begin{array}{cc}1& 0\\ 2& -4\end{array}\right]$, verify that
(i) (2A)^T = 2A^T
(ii) (A + B)^T = A^T + B^T
(iii) (A − B)^T = A^T − B^T
(iv) (AB)^T = B^T A^T

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}A=\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right] \\ {A}^T=\left[\begin{array}{cc}2& -7\\ -3& 5\end{array}\right] \\ B=\left[\begin{array}{cc}1& 0\\ 2& -4\end{array}\right] \\ {B}^T=\left[\begin{array}{cc}1& 2\\ 0& -4\end{array}\right] \\ \left(i\right) \\ {\left(2A\right)}^T=2A^T \\ \Rightarrow {\left(2\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right]\right)}^T=2\left[\begin{array}{cc}2& -7\\ -3& 5\end{array}\right] \\ \Rightarrow {\left[\begin{array}{cc}4& -6\\ -14& 10\end{array}\right]}^T=\left[\begin{array}{cc}4& -14\\ -6& 10\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}4& -14\\ -6& 10\end{array}\right]=\left[\begin{array}{cc}4& -14\\ -6& 10\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \left(ii\right){\left(A+B\right)}^T=A^T+B^T \\ \Rightarrow {\left(\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 2& -4\end{array}\right]\right)}^T=\left[\begin{array}{cc}2& -7\\ -3& 5\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 0& -4\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}2+1& -3+0\\ -7+2& 5-4\end{array}\right]\right)}^T=\left[\begin{array}{cc}2+1& -7+2\\ -3+0& 5-4\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}3& -3\\ -5& 1\end{array}\right]\right)}^T=\left[\begin{array}{cc}3& -5\\ -3& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}3& -5\\ -3& 1\end{array}\right]=\left[\begin{array}{cc}3& -5\\ -3& 1\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \left(iii\right){\left(A-B\right)}^T=A^T-B^T \\ \Rightarrow {\left(\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 2& -4\end{array}\right]\right)}^T=\left[\begin{array}{cc}2& -7\\ -3& 5\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& -4\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}2-1& -3-0\\ -7-2& 5+4\end{array}\right]\right)}^T=\left[\begin{array}{cc}2-1& -7-2\\ -3-0& 5+4\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}1& -3\\ -9& 9\end{array}\right]\right)}^T=\left[\begin{array}{cc}1& -9\\ -3& 9\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}1& -9\\ -3& 9\end{array}\right]=\left[\begin{array}{cc}1& -9\\ -3& 9\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \left(iv\right){\left(AB\right)}^T=B^T{A}^T \\ \Rightarrow {\left(\left[\begin{array}{cc}2& -3\\ -7& 5\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& -4\end{array}\right]\right)}^T=\left[\begin{array}{cc}1& 2\\ 0& -4\end{array}\right]\left[\begin{array}{cc}2& -7\\ -3& 5\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}2-6& 0+12\\ -7+10& 0-20\end{array}\right]\right)}^T=\left[\begin{array}{cc}2-6& -7+10\\ 0+12& 0-20\end{array}\right] \\ \Rightarrow {\left(\left[\begin{array}{cc}-4& 12\\ 3& -20\end{array}\right]\right)}^T=\left[\begin{array}{cc}-4& 3\\ 12& -20\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}-4& 3\\ 12& -20\end{array}\right]=\left[\begin{array}{cc}-4& 3\\ 12& -20\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$