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 \[{(AB)^T}\] is
equal to
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. \[\left[
{\begin{array}{*{20}{c}}{ - 3}&7\\{10}&2\end{array}} \right]\]
D. None of these

Answer 1

Hint:

First of all we are given two matrices which are 2×3 and 3×2 and we have to find its product AB and then its transpose. So when we multiply these two matrices we will get a 2×2 matrix.


Formula Used:
Let a matrix A= \[\left[{\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\end{array}}\right]\] and another matrix B= \[\left[{\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}\\{{b_{21}}}&{{b_{22}}}\\{{b_{31}}}&{{b_{32}}}\end{array}}\right]\]
Now as we multiply these two matrices as we will multiply rows of first matrix with a column of the second matrix we will get a final matrix as 2×2 matrix
So , AB= \[\left[{\begin{array}{*{20}{c}}{{a_{11}}{b_{11}} + {a_{12}}{b_{21}} + {a_{13}}{b_{31}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}} + {a_{13}}{b_{32}}}\\{{a_{21}}{b_{11}} + {a_{22}}{b_{21}} + {a_{23}}{b_{31}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{12}} + {a_{23}}{b_{32}}}\end{array}} \right]\]
Complete Step-By-Step Solution:
We are given a matrix A= \[\left[{\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right]\]and
another matrix B= \[\left[{\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right]\]
Now we have to find \[{(AB)^T}\].
So for that first of all we have to find its product of AB
So AB = \[\left[ {\begin{array}{*{20}{c}}{1(2) + ( - 2)(3) +1(1)}&{1(1) + ( - 2)(2) + 1(1)}\\{2(2) + (1)(3) + 3(1)}&{2(1) + (1)(2)+ 3(1)}\end{array}} \right]\]
\[AB = \left[{\begin{array}{*{20}{c}}{2 - 6 + 1}&{1 - 4 + 1}\\{4 + 3 + 3}&{2 + 2 +3}\end{array}} \right]\]
\[AB = \left[{\begin{array}{*{20}{c}}{ - 3}&{ - 2}\\{10}&7\end{array}} \right]\]
Now as we have find AB now we have to find its transpose
\[{(AB)^T} = \left[{\begin{array}{*{20}{c}}{ - 3}&{10}\\{ - 2}&7\end{array}} \right]\]
As in transpose, we have to interchange rows to columns and columns
to rows.
Option B is correct.

Note:
for multiplication of 2×3 and 3×2 matrix, we get 2×2 matrix so here is the process how we get it
A . B = AB
m×n n×p m×p
as n and n are equal so they cancel each other and we get m×p as the required matrix students should remember while solving such questions as the require knowledge of matrix multiplication and which matrix is of which form
In general, matrix multiplication, unlike arithmetic multiplication, is not commutative, which means the multiplication of matrix A and B, given as AB, cannot be equal to BA, i.e., AB ≠ BA.
Therefore, the order of multiplication for the multiplication of matrices is important.