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”.