Chapter 8, Problem 6PS

To determine

Find the transpose of the given matrices and verify that the given expression.

Answer to Problem 6PS

The transpose of the given matrices are

  $\begin{array}{l}{A}^T=\left[\begin{array}{cc}-1& 2\\ 1& 0\\ -2& 1\end{array}\right]\\ B^T=\left[\begin{array}{ccc}-3& 1& 1\\ 0& 2& -1\end{array}\right]\end{array}$

Explanation of Solution

Given:

The given matrices are $A=\left[\begin{array}{ccc}-1& 1& -2\\ 2& 0& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}-3& 0\\ 1& 2\\ 1& -1\end{array}\right]$ .

Calculation:

The transpose of a matrix is formed by writing the columns of the matrix as rows.

  $\begin{array}{l}{A}^T=\left[\begin{array}{cc}-1& 2\\ 1& 0\\ -2& 1\end{array}\right]\\ B^T=\left[\begin{array}{ccc}-3& 1& 1\\ 0& 2& -1\end{array}\right]\end{array}$

The given expression is ${\left(AB\right)}^T=B^T{A}^T$

L.H.S is ${\left(AB\right)}^T$

  $\begin{array}{l}AB=\left[\begin{array}{ccc}-1& 1& -2\\ 2& 0& 1\end{array}\right]\left[\begin{array}{cc}-3& 0\\ 1& 2\\ 1& -1\end{array}\right]\\ =\left[\begin{array}{cc}-1\cdot \left(-3\right)+1\cdot 1+\left(-2\right)1& -1\cdot 0+1\cdot 2+\left(-2\right)\left(-1\right)\\ 2\cdot \left(-3\right)+0\cdot 1+1\cdot 1& 2\cdot 0+0\cdot 2+1\cdot \left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}2& 4\\ -5& -1\end{array}\right]\end{array}$

The transpose of a matrix is formed by writing the columns of the matrix as rows.

  ${\left(AB\right)}^T=\left[\begin{array}{cc}2& -5\\ 4& -1\end{array}\right]$

R.H.S is $B^T{A}^T$ .

  $\begin{array}{l}{B}^T{A}^T=\left[\begin{array}{ccc}-3& 1& 1\\ 0& 2& -1\end{array}\right]\left[\begin{array}{cc}-1& 2\\ 1& 0\\ -2& 1\end{array}\right]\\ =\left[\begin{array}{cc}\left(-3\right)\left(-1\right)+1\cdot 1+1\left(-2\right)& \left(-3\right)\cdot 2+1\cdot 0+1\cdot 1\\ 0\left(-1\right)+2\cdot 1+\left(-1\right)\cdot \left(-2\right)& 0\cdot 2+2\cdot 0+\left(-1\right)\cdot 1\end{array}\right]\\ =\left[\begin{array}{cc}2& -5\\ 4& -1\end{array}\right]\end{array}$

Hence L.H.S $=$ R.H.S proved.