Question

If $A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ and $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$, then find ${\left( {AB} \right)^T}$.
A. $\left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  {10}&7
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
  { - 3}&{10} \\
  { - 2}&7
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  { - 3}&{10} \\
  7&{ - 2}
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right]$

Answer 1

Hint: In this question we have to find the product of the two given matrices, A and B. Then, we have to find the transpose of the resultant matrix. To solve these types of questions, we must know the basic operations on matrices, such as addition, subtraction, multiplication, and transpose.

Complete step by step Solution:
Given: $\;A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ and $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$
For multiplication of matrix:
Step 1: Multiply and add each element of the first row of matrix $A$with the respective elements of the first column of matrix $B$, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$
Result for this will be, $R1 = \;\left[ {(1 \times 2) + (( - 2) \times 3) + (1 \times 1)} \right]$
Step 2: Multiply and add each element of the first row of matrix A with the respective elements of the second column of matrix B, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$
Result for this will be, $R2 = \left[ {(1 \times 1) + (( - 2) \times 2) + (1 \times 1)} \right]$
Step 3: Multiply and add each element of the second row of matrix A with the respective elements of the first column of matrix B, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$
Result for this will be, $R3 = \left[ {(2 \times 2) + (1 \times 3) + (3 \times 1)} \right]$
Step 4: Multiply and add each element of the second row of matrix A with the respective elements of the second column of matrix B, as shown below:
$A = \;\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&1 \\
  2&1&3
\end{array}} \right]$ $B = \;\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2 \\
  1&1
\end{array}} \right]$
Result for this will be, $R4 = \left[ {(2 \times 1) + (1 \times 2) + (3 \times 1)} \right]$
Step 5: Matrix multiplication of matrices A and B will be,
$AB = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R2} \\
  {R3}&{R4}
\end{array}} \right]$
Now, substituting the values of $R1,R2,R3$ and \[R4,\] we get:
$AB = \left[ {\begin{array}{*{20}{c}}
  {(1 \times 2) + (( - 2) \times 3) + (1 \times 1)}&{(1 \times 1) + (( - 2) \times 2) + (1 \times 1)} \\
  {(2 \times 2) + (1 \times 3) + (3 \times 1)}&{(2 \times 1) + (1 \times 2) + (3 \times 1)}
\end{array}} \right]$
After solving the above expressions, we get:
$AB = \left[ {\begin{array}{*{20}{c}}
  {2 + \left( { - 6} \right) + \;1}&{\;\;\;\;\;\;1 + \left( { - 4} \right) + 1} \\
  {4 + 3 + 3}&{2 + 2 + 3}
\end{array}} \right]$
$AB = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  {10}&7
\end{array}} \right]$

Now, for transposing the matrix:
Step 1: Interchange the rows with columns, as shown below:
${\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  {R1}&{R3} \\
  {R2}&{R4}
\end{array}} \right]$
Step 2: Final transpose matrix, ${\text{=}}{\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{10} \\
  { - 2}&7
\end{array}} \right]$

Therefore, the correct option is (B).

Additional Information:In this type of question, where we have to multiply two given matrices, whose order is different, like a matrix A of order \[\left( {m \times n} \right)\], where ‘m’ is the number of rows of matrix A and ‘n’ is the number of columns of matrix A.
Another matrix B of order \[\left( {n \times p} \right)\], where ‘n’ is the number of rows of matrix B and ‘p’ is the number of columns of matrix B.
Then, the order of the resultant matrix should be\[\left( {m \times p} \right)\].

Note: In this type of question, where we have to multiply two given matrices, if the number of columns of the first matrix is equal to the number of rows of the second matrix, then multiplication is possible. Otherwise, multiplication is not possible.