Question

If $A = \left[ {\begin{array}{*{20}{c}}
  { - 1} \\
  2 \\
  3
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 1}&{ - 4}
\end{array}} \right]$, verify that ${\left( {AB} \right)^T} = {B^T}{A^T}$.

Answer 1

Hint: First we need to find the product of the 2 matrices that are provided as A and B. Then by taking the transpose of the produce we will get ${\left( {AB} \right)^T}$. Again we can find the individual transpose matrices of A and B and find their product. We will see that the result is the same in both cases.
Formula used:
In this formula we will be using the following formula,
If $A = \left[ {\begin{array}{*{20}{c}}
  {{A_1}} \\
  {{A_2}} \\
  {{A_3}}
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  {{B_1}}&{{B_2}}&{{B_3}}
\end{array}} \right]$
then $AB = \left[ {\begin{array}{*{20}{c}}
  {{A_1}} \\
  {{A_2}} \\
  {{A_3}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {{B_1}}&{{B_2}}&{{B_3}}
\end{array}} \right]$
that is, $AB = \left[ {\begin{array}{*{20}{c}}
  {{A_1}{B_1}}&{{A_1}{B_2}}&{{A_1}{B_3}} \\
  {{A_2}{B_1}}&{{A_2}{B_2}}&{{A_2}{B_3}} \\
  {{A_3}{B_1}}&{{A_3}{B_2}}&{{A_3}{B_3}}
\end{array}} \right]$

Complete step by step answer:
We need to first calculate the product of the 2 matrices A and B
So we use the formula $AB = \left[ {\begin{array}{*{20}{c}}
  {{A_1}{B_1}}&{{A_1}{B_2}}&{{A_1}{B_3}} \\
  {{A_2}{B_1}}&{{A_2}{B_2}}&{{A_2}{B_3}} \\
  {{A_3}{B_1}}&{{A_3}{B_2}}&{{A_3}{B_3}}
\end{array}} \right]$ to calculate the product.
In the question we are given $A = \left[ {\begin{array}{*{20}{c}}
  { - 1} \\
  2 \\
  3
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 1}&{ - 4}
\end{array}} \right]$. So substituting the values we get the matrix product as,
\[AB = \left[ {\begin{array}{*{20}{c}}
  { - 1 \times - 2}&{ - 1 \times - 1}&{ - 1 \times - 4} \\
  {2 \times - 2}&{2 \times - 1}&{2 \times - 4} \\
  {3 \times - 2}&{3 \times - 1}&{3 \times - 4}
\end{array}} \right]\]
So on calculating we get the product as,
\[AB = \left[ {\begin{array}{*{20}{c}}
  2&1&4 \\
  { - 4}&{ - 2}&{ - 8} \\
  { - 6}&{ - 3}&{ - 12}
\end{array}} \right]\]
We can get the transpose of this matrix as,
\[{\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  2&{ - 4}&{ - 6} \\
  1&{ - 2}&{ - 3} \\
  4&{ - 8}&{ - 12}
\end{array}} \right]\]. Let us name this as equation1.
Now from A we get ${A^T} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&2&3
\end{array}} \right]$
and similarly from B we get, ${B^T} = \left[ {\begin{array}{*{20}{c}}
  { - 2} \\
  { - 1} \\
  { - 4}
\end{array}} \right]$
Therefore, the product of the transpose of the A matrix with the transpose of the B matrix gives us,
${B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  { - 2} \\
  { - 1} \\
  { - 4}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 1}&2&3
\end{array}} \right]$Again from the same formula for the product of matrices, we get
\[{B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  { - 2 \times - 1}&{ - 2 \times 2}&{ - 2 \times 3} \\
  { - 1 \times - 1}&{ - 1 \times 2}&{ - 1 \times 3} \\
  { - 4 \times - 1}&{ - 4 \times 2}&{ - 4 \times 3}
\end{array}} \right]\]
So on doing the multiplications we get,
\[{B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  2&{ - 4}&{ - 6} \\
  1&{ - 2}&{ - 3} \\
  4&{ - 8}&{ - 12}
\end{array}} \right]\]
Let us consider this as equation 2.
So on comparing the values in equation 1 and 2 we can see that the values in the RHS of both the equations are the same. Hence we get,
${\left( {AB} \right)^T} = {B^T}{A^T}$

Note:
The transpose of a matrix is obtained by interchanging the rows and the columns of a matrix. In other words, we can denote $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$ matrix in the form of $A = {\left[ {{a_{ji}}} \right]_{n \times m}}$.