Question

If $A = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&1&2 \\
  2&3&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  { - 5}&7&1 \\
  1&{ - 5}&7 \\
  7&1&{ - 5}
\end{array}} \right)$ then $AB$ is equal to
Option:
A. ${I_3}$
B. $2{I_3}$
C. $4{I_3}$
D. $18{I_3}$

Answer 1

Hint: If the number of columns in the first matrix equals the number of rows in the second matrix, the product of the two matrices will be known. The resultant matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix if the product is defined.

Formula Used: The product $AB$ is an $m \times p$ matrix if $A = \left[ {{a_{ij}}} \right]$ is an $m \times n$ matrix and $B = \left[ {{b_{ij}}} \right]$ is an $n \times p$ matrix.
$AB = \left[ {{c_{ij}}} \right]$ , where $\left[ {{c_{ij}}} \right]$ is composed of ${a_{1j}}{b_{1j}} + {a_{2j}}{b_{2j}} + \ldots + {a_{in}}{b_{nj}}$ .

Complete step by step solution: We have matrices $A = \left( {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&1&2 \\
  2&3&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  { - 5}&7&1 \\
  1&{ - 5}&7 \\
  7&1&{ - 5}
\end{array}} \right)$ .
To find the product of these two matrices, we have to multiply each row of one matrix to each column of the other matrix respectively.
$AB = \left( {\begin{array}{*{20}{c}}
  {1 \times ( - 5) + 2 \times 1 + 3 \times 7}&{3 \times ( - 5) + 1 \times 1 + 2 \times 7}&{2 \times ( - 5) + 3 \times 1 + 1 \times 7} \\
  {1 \times 7 + 2 \times ( - 5) + 3 \times 1}&{3 \times 7 + 1 \times ( - 5) + 2 \times 1}&{2 \times 7 + 3 \times ( - 5) + 1 \times 1} \\
  {1 \times 1 + 2 \times 7 + 3 \times ( - 5)}&{3 \times 1 + 1 \times 7 + 2 \times ( - 5)}&{2 \times 1 + 3 \times 7 + 1 \times ( - 5)}
\end{array}} \right)$
On solving, we get
$AB = \left( {\begin{array}{*{20}{c}}
  {18}&0&0 \\
  0&{18}&0 \\
  0&0&{18}
\end{array}} \right)$
$ \Rightarrow AB = 18{I_3}$

Option ‘D’ is correct

Note: To multiply two matrices, we must ensure that the number of rows in the second matrix equals the number of columns in the first matrix. As a result, a specific number of rows from the first matrix and a specific number of columns from the second matrix will be present in the final matrix. The matrix multiplication order determines the order of the final matrix.