Question

If A = \left( {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 5&7&9 \\\ { - 2}&1&1 \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} { - 4}&1&{ - 5} \\\ 1&2&0 \\\ 1&3&1 \end{array}} \right) , then verify that
${\left( {A + B} \right)^T} = {A^T} + {B^T}$
${\left( {A - B} \right)^T} = {A^T} - {B^T}$

Answer 1

Hint: Here ${X^T}$means the transpose of the matrix $X$. First find the transposes of the matrix and then solve accordingly in the problem by comparing the LHS and RHS.

Given that,
A = \left( {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 5&7&9 \\\ { - 2}&1&1 \end{array}} \right) and B = \left( {\begin{array}{*{20}{c}} { - 4}&1&{ - 5} \\\ 1&2&0 \\\ 1&3&1 \end{array}} \right)
Consider $A + B$
A + B = \left( {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 5&7&9 \\\ { - 2}&1&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 4}&1&{ - 5} \\\ 1&2&0 \\\ 1&3&1 \end{array}} \right) \\\ \\\ A + B = \left( {\begin{array}{*{20}{c}} { - 1 - 4}&{2 + 1}&{3 - 5} \\\ {5 + 1}&{7 + 2}&{9 + 0} \\\ { - 2 + 1}&{1 + 3}&{1 + 1} \end{array}} \right) \\\ \\\ A + B = \left( {\begin{array}{*{20}{c}} { - 5}&3&{ - 2} \\\ 6&9&9 \\\ { - 1}&4&2 \end{array}} \right) \\\
Now consider $A - B$
A - B = \left( {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 5&7&9 \\\ { - 2}&1&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 4}&1&{ - 5} \\\ 1&2&0 \\\ 1&3&1 \end{array}} \right) \\\ \\\ A - B = \left( {\begin{array}{*{20}{c}} { - 1 - ( - 4)}&{2 - 1}&{3 - ( - 5)} \\\ {5 - 1}&{7 - 2}&{9 - 0} \\\ { - 2 - 1}&{1 - 3}&{1 - 1} \end{array}} \right) \\\ \\\ A - B = \left( {\begin{array}{*{20}{c}} 3&1&8 \\\ 4&5&9 \\\ { - 3}&{ - 2}&0 \end{array}} \right) \\\
Consider the transpose of $(A + B)$ i.e. ${(A + B)^T}$
{(A + B)^T} = {\left( {\begin{array}{*{20}{c}} { - 5}&3&{ - 2} \\\ 6&9&9 \\\ { - 1}&4&2 \end{array}} \right)^T} \\\ \\\ {(A + B)^T} = \left( {\begin{array}{*{20}{c}} { - 5}&6&{ - 1} \\\ 3&9&4 \\\ { - 2}&9&2 \end{array}} \right) \\\ \\\ \therefore {(A + B)^T} = \left( {\begin{array}{*{20}{c}} { - 5}&6&{ - 1} \\\ 3&9&4 \\\ { - 2}&9&2 \end{array}} \right)......................\left( 1 \right) \\\
Now consider the transpose of $(A - B)$ i.e. ${(A - B)^T}$
{(A - B)^T} ={ \left( {\begin{array}{*{20}{c}} 3&1&8 \\\ 4&5&9 \\\ { - 3}&{ - 2}&0 \end{array}} \right)^T} \\\ \\\ {\left( {A - B} \right)^T} = \left( {\begin{array}{*{20}{c}} 3&4&{ - 3} \\\ 1&5&{ - 2} \\\ 8&9&0 \end{array}} \right) \\\ \\\ \therefore {\left( {A - B} \right)^T} = \left( {\begin{array}{*{20}{c}} 3&4&{ - 3} \\\ 1&5&{ - 2} \\\ 8&9&0 \end{array}} \right)........................\left( 2 \right) \\\
In the same way find transpose of $A$ i.e. ${A^T}$

{A^T} = {\left( {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 5&7&9 \\\ { - 2}&1&1 \end{array}} \right)^T} \\\ \\\ \therefore {A^T} = \left( {\begin{array}{*{20}{c}} { - 1}&5&{ - 2} \\\ 2&7&1 \\\ 3&9&1 \end{array}} \right) \\\

Now similarly the transpose of $B$ i.e. ${B^T}$
{B^T} = {\left( {\begin{array}{*{20}{c}} { - 4}&1&{ - 5} \\\ 1&2&0 \\\ 1&3&1 \end{array}} \right)^T} \\\ \\\ \therefore {B^T} = \left( {\begin{array}{*{20}{c}} { - 4}&1&1 \\\ 1&2&3 \\\ { - 5}&0&1 \end{array}} \right) \\\
Now find ${A^T} + {B^T}$ i.e.

{A^T} + {B^T} = \left( {\begin{array}{*{20}{c}} { - 1}&5&{ - 2} \\\ 2&7&1 \\\ 3&9&1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 4}&1&1 \\\ 1&2&3 \\\ { - 5}&0&1 \end{array}} \right) \\\ \\\ {A^T} + {B^T} = \left( {\begin{array}{*{20}{c}} { - 1 - 4}&{5 + 1}&{ - 2 + 1} \\\ {2 + 1}&{7 + 2}&{1 + 3} \\\ {3 - 5}&{9 + 0}&{1 + 1} \end{array}} \right) \\\ \\\ \therefore {A^T} + {B^T} = \left( {\begin{array}{*{20}{c}} { - 5}&6&{ - 1} \\\ 3&9&4 \\\ { - 2}&9&2 \end{array}} \right)..................................\left( 3 \right) \\\

Now find ${A^T} - {B^T}$ i.e.

{A^T} - {B^T} = \left( {\begin{array}{*{20}{c}} { - 1}&5&{ - 2} \\\ 2&7&1 \\\ 3&9&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 4}&1&1 \\\ 1&2&3 \\\ { - 5}&0&1 \end{array}} \right) \\\ \\\ {A^T} - {B^T} = \left( {\begin{array}{*{20}{c}} { - 1 + 4}&{5 - 1}&{ - 2 - 1} \\\ {2 - 1}&{7 - 2}&{1 - 3} \\\ {3 + 5}&{9 - 0}&{1 - 1} \end{array}} \right) \\\ \\\ \therefore {A^T} - {B^T} = \left( {\begin{array}{*{20}{c}} 3&4&{ - 3} \\\ 1&5&{ - 2} \\\ 8&9&0 \end{array}} \right)................................................\left( 4 \right) \\\

From Equation $\left( 1 \right)$and Equation $\left( 3 \right)$we have
${\left( {A + B} \right)^T} = {A^T} + {B^T}$
From Equation $\left( 2 \right)$and Equation $\left( 4 \right)$ we have
${\left( {A - B} \right)^T} = {A^T} - {B^T}$
Hence proved that ${\left( {A + B} \right)^T} = {A^T} + {B^T}$
${\left( {A - B} \right)^T} = {A^T} - {B^T}$

Note: From this problem it is clear that ${\left( {A + B} \right)^T} = {A^T} + {B^T}$is the property of “Transpose of a sum of matrices” and ${\left( {A - B} \right)^T} = {A^T} - {B^T}$is the property of “Transpose of subtraction of matrices”.