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
363.6k+ views
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”.
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”.
Last updated date: 19th Sep 2023
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