Question

If $A = \left[ {\begin{array}{*{20}{c}}
  0&2&0 \\
  0&0&3 \\
  { - 2}&2&0
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&4&5 \\
  5&{ - 4}&0
\end{array}} \right]$ , then the element of third row and third column in $AB$ will be
A. $ - 18$
B. $4$
C. $ - 12$
D. None of these

Answer 1

Hint: The first matrix's columns must have the same number of rows as the second matrix's rows in order for matrices to be multiplied. The final matrix, also known as the matrix product, is created by adding the number of rows in the first matrix and the number of columns in the second matrix. The outcome of the matrices A and B is represented by the letters AB.

Formula Used: Let a matrix $A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]$ , $B = \left[ {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\
  {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \\
  {{b_{31}}}&{{b_{32}}}&{{b_{33}}}
\end{array}} \right]$ then their product will be: $AB = A \cdot B = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}{b_{11}} + {a_{12}}{b_{21}} + {a_{13}}{b_{31}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}} + {a_{13}}{b_{32}}}&{{a_{11}}{b_{13}} + {a_{12}}{b_{23}} + {a_{13}}{b_{33}}} \\
  {{a_{21}}{b_{11}} + {a_{22}}{b_{21}} + {a_{23}}{b_{31}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{22}} + {a_{23}}{b_{32}}}&{{a_{21}}{b_{13}} + {a_{22}}{b_{23}} + {a_{23}}{b_{33}}} \\
  {{a_{31}}{b_{11}} + {a_{32}}{b_{21}} + {a_{33}}{b_{31}}}&{{a_{31}}{b_{12}} + {a_{32}}{b_{22}} + {a_{33}}{b_{32}}}&{{a_{31}}{b_{13}} + {a_{32}}{b_{23}} + {a_{33}}{b_{33}}}
\end{array}} \right]$ .

Complete step by step Solution:
We have matrices $A = \left[ {\begin{array}{*{20}{c}}
  0&2&0 \\
  0&0&3 \\
  { - 2}&2&0
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  3&4&5 \\
  5&{ - 4}&0
\end{array}} \right]$ .
Now, to find the product of two matrices applies the formula:
$AB = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}{b_{11}} + {a_{12}}{b_{21}} + {a_{13}}{b_{31}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}} + {a_{13}}{b_{32}}}&{{a_{11}}{b_{13}} + {a_{12}}{b_{23}} + {a_{13}}{b_{33}}} \\
  {{a_{21}}{b_{11}} + {a_{22}}{b_{21}} + {a_{23}}{b_{31}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{22}} + {a_{23}}{b_{32}}}&{{a_{21}}{b_{13}} + {a_{22}}{b_{23}} + {a_{23}}{b_{33}}} \\
  {{a_{31}}{b_{11}} + {a_{32}}{b_{21}} + {a_{33}}{b_{31}}}&{{a_{31}}{b_{12}} + {a_{32}}{b_{22}} + {a_{33}}{b_{32}}}&{{a_{31}}{b_{13}} + {a_{32}}{b_{23}} + {a_{33}}{b_{33}}}
\end{array}} \right]$
Substituting the $A$ and $B$matrices elements into this formula to get the product $AB$, we will have
$AB = \left[ {\begin{array}{*{20}{c}}
  {0 \times 1 + 2 \times 3 + 0 \times 5}&{0 \times 2 + 2 \times 4 + 0 \times ( - 4)}&{0 \times 3 + 2 \times 5 + 0 \times 0} \\
  {0 \times 1 + 0 \times 3 + 3 \times 5}&{0 \times 2 + 0 \times 4 + 3 \times ( - 4)}&{0 \times 3 + 0 \times 5 + 3 \times 0} \\
  {( - 2) \times 1 + 2 \times 3 + 0 \times 5}&{( - 2) \times 2 + 2 \times 4 + 0 \times ( - 4)}&{( - 2) \times 3 + 2 \times 5 + 0 \times 0}
\end{array}} \right]$
On solving the addition in the matrix elements, we get
$AB = \left[ {\begin{array}{*{20}{c}}
  6&8&{10} \\
  {15}&{ - 12}&0 \\
  4&4&4
\end{array}} \right]$
Hence, we enter the entry in the third row and third column as $({a_{31}}{b_{13}} + {a_{32}}{b_{23}} + {a_{33}}{b_{33}})$ in matrix $AB$ i.e., $4$ .

Therefore, the correct option is (B).

Note: When multiplying two matrices together, we shall use the dot product. The rows and columns of a matrix should be treated as a vector when it is multiplied by another matrix. We want to treat every row in the first matrix as a vector and every column in the second matrix as a vector, to be more precise.