Question

Which of the following is not true, if $A$ and $B$ are two matrices each of order $n \times n$ then
A.${\left( {A + B} \right)^T} = {B^T} + {A^T}$
B.${\left( {A - B} \right)^T} = {A^T} - {B^T}$
C.${\left( {AB} \right)^T} = {A^T}{B^T}$
D.${\left( {ABC} \right)^T} = {C^T}{B^T}{A^T}$

Answer 1

The given question is about the matrices. Matrices are a tabular method or tabular way to represent the data. Further subtraction, addition, multiplication and division are also possible in matrices. It means matrices hold Binary operations. In the question we had asked that out of $4$ options which one is correct.

Complete step-by-step answer:
In the given question we had to check four options about matrices. For them, let

1&2 \\\ 3&4 \end{array}} \right]{\text{ and }}B = \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]{\text{ and }}C = \left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]{\text{ }}$$ and ${A^t}$ represent transpose means rows changes to columns and vice versa In option (i) ${\left( {A - B} \right)^t} = {B^t} - {A^t}$ Putting here values $${\left( {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]} \right)^t} = {\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]^t} - {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]^t}$$ On taking transpose $${\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\\ { - 1}&3 \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 3&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]$$ On subtraction, $$\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\\ { - 1}&3 \end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}} 1&1 \\\ 1&{ - 3} \end{array}} \right]$$ Which is not equal Therefore option (A) is incorrect In option (ii) ${\left( {A - B} \right)^t} = {A^t} - {B^t}$ Putting here the values of $A{\text{ and }}B$ we get $${\left( {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]} \right)^t} = {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]^t} - {\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]^t}$$ Subtracting and taking transpose on left side $${\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\\ { - 1}&3 \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 3&1 \end{array}} \right]$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&3 \end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\\ { - 1}&3 \end{array}} \right]$$ Which is also not equal So option (ii) is also incorrect In option (iii) ${\left( {AB} \right)^t} = {A^t}{B^t}$ Putting here values of$A{\text{ \& }}B$, we get ${\left( {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]} \right)^t} = {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]^t}{\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]^t}$ Let left side, do the multiply. This can be done by multiplying rows of first and columns of second and then adding, we get ${\left[ {\begin{array}{*{20}{c}} {2 + 8}&{3 + 2} \\\ {6 + 16}&{9 + 4} \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]{\text{ }}\left[ {\begin{array}{*{20}{c}} 2&4 \\\ 3&1 \end{array}} \right]$ $ \Rightarrow {\left[ {\begin{array}{*{20}{c}} {10}&5 \\\ {22}&{13} \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} {2 + 3}&{4 + 3} \\\ {4 + 12}&{8 + 4} \end{array}} \right]{\text{ }}$ Which is also not equal So, option (iii) is also incorrect In option $${\left( {ABC} \right)^t} = {C^t}{B^t}{A^t}$$ Substituting here the values of $A,B{\text{ and }}C$, we get $${\left( {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]} \right)^t} = {\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]^t}{\left[ {\begin{array}{*{20}{c}} 2&3 \\\ 4&1 \end{array}} \right]^t}{\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]^t}$$ On left side, firstly multiply first two matrices and then multiply the resultant by third one and taking transfer, we get $${\left( {\left[ {\begin{array}{*{20}{c}} {2 + 8}&{3 + 2} \\\ {6 + 16}&{9 + 4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]} \right)^t} = \left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&4 \\\ 3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]$$ It eight side, taking transpose and then do the multiply same as in left side. ${\left( {\left[ {\begin{array}{*{20}{c}} {10}&5 \\\ {22}&{13} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]} \right)^t} = \left[ {\begin{array}{*{20}{c}} {2 + 6}&{4 + 2} \\\ {6 + 12}&{12 + 4} \end{array}} \right]{\text{ }}\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]$ Which further on multiplication given $${\left[ {\begin{array}{*{20}{c}} {10 + 10}&{30 + 20} \\\ {22 + 26}&{66 + 52} \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} 8&6 \\\ {18}&{16} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 2&4 \end{array}} \right]$$ $$ \Rightarrow {\left[ {\begin{array}{*{20}{c}} {20}&{50} \\\ {48}&{118} \end{array}} \right]^t} = \left[ {\begin{array}{*{20}{c}} {8 + 12}&{24 + 24} \\\ {18 + 32}&{54 + 64} \end{array}} \right]$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {20}&{50} \\\ {48}&{118} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {20}&{48} \\\ {50}&{118} \end{array}} \right]$$ Which holds true Which means option (iv) is correct. **Hence, option A,B and C is not true for matrices.** **Note:** This property of matrices hold true even if we took two matrices and tens property of matrices ${\left( {ABC} \right)^t} = {C^t}{B^t}{A^t}$ implies ${\left( {AB} \right)^t} = {B^t}{A^t}$, which holds true for two matrices $A{\text{ and }}B$ instead of three matrices $A,B,C$. Also, this property holds true for any order of matrices.