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# Find the value of X and Y if$2X + 3Y = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right]$ and $3X + 2Y = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ { - 1}&5 \end{array}} \right]$

Last updated date: 23rd Jun 2024
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Hint:Here first we will assume the two given matrices to be P and Q respectively and then form two linear equations.We will then solve those equations by elimination method and find the values of X and Y.

The given equations are:
$2X + 3Y = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right]$ and $3X + 2Y = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ { - 1}&5 \end{array}} \right]$
Let us assume
$\left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right] = P$ and $\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ { - 1}&5 \end{array}} \right] = Q$
Therefore putting these values in the above equations we get:-
$2X + 3Y = P...................\left( 1 \right) \\ 3X + 2Y = Q...............\left( 2 \right) \\$
Now we will solve these equations using the elimination method.
Hence multiplying equation 1 by 3 we get:-
$3 \times \left( {2X + 3Y = P} \right)$
Solving it further we get:-
$3 \times 2X + 3 \times 3Y = 3 \times P \\ 6X + 9Y = 3P....................\left( 3 \right) \\$
Now multiplying equation 2 by 2 we get:-
$2 \times \left( {3X + 2Y = Q} \right)$
Solving it further we get:-
$2 \times 3X + 2 \times 2Y = 2 \times Q \\ 6X + 4Y = 2Q........................\left( 4 \right) \\$
Now subtracting equation 4 from equation 3 we get:-
$6X + 9Y = 3P \\ 6X + 4Y = 2Q \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ {\text{ }}5Y = 3P - 2Q \\$
Now evaluating the value of y we get:-
$Y = \dfrac{{3P - 2Q}}{5} \\ \Rightarrow Y = \dfrac{3}{5}P - \dfrac{2}{5}Q \\$
Now putting in the values of P and Q we get:-
$Y = \dfrac{3}{5}\left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right] - \dfrac{2}{5}\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ { - 1}&5 \end{array}} \right]$
Now multiplying and solving it further we get:-
$Y = \left[ {\begin{array}{*{20}{c}} {\dfrac{6}{5}}&{\dfrac{9}{5}} \\ {\dfrac{{12}}{5}}&0 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {\dfrac{4}{5}}&{\dfrac{{ - 4}}{5}} \\ {\dfrac{{ - 2}}{5}}&{\dfrac{{10}}{5}} \end{array}} \right]$
Solving it further we get:-
$\Rightarrow Y = \left[ {\begin{array}{*{20}{c}} {\dfrac{6}{5} - \dfrac{4}{5}}&{\dfrac{9}{5} - \left( {\dfrac{{ - 4}}{5}} \right)} \\ {\dfrac{{12}}{5} - \left( {\dfrac{{ - 2}}{5}} \right)}&{0 - \left( {\dfrac{{10}}{5}} \right)} \end{array}} \right] \\ \Rightarrow Y = \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{13}}{5}} \\ {\dfrac{{14}}{5}}&{\dfrac{{ - 10}}{5}} \end{array}} \right] \\$
Now putting this value in equation 1 we get:-
The equation 1 is given by:-
$2X + 3Y = P$
Now putting the known values of Y and P we get:-
$2X + 3\left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{13}}{5}} \\ {\dfrac{{14}}{5}}&{\dfrac{{ - 10}}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right]$
Now on multiplying by 3 we get:-
$2X + \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5} \times 3}&{\dfrac{{13}}{5} \times 3} \\ {\dfrac{{14}}{5} \times 3}&{\dfrac{{ - 10}}{5} \times 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right] \\ 2X + \left[ {\begin{array}{*{20}{c}} {\dfrac{6}{5}}&{\dfrac{{39}}{5}} \\ {\dfrac{{42}}{5}}&{\dfrac{{ - 30}}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right] \\$
Now simplifying for the value of X we get:-
$2X = \left[ {\begin{array}{*{20}{c}} {2 - \dfrac{6}{5}}&{3 - \dfrac{{39}}{5}} \\ {4 - \dfrac{{42}}{5}}&{0 - \dfrac{{30}}{5}} \end{array}} \right] \\ 2X = \left[ {\begin{array}{*{20}{c}} {\dfrac{{10 - 6}}{5}}&{\dfrac{{15 - 39}}{5}} \\ {\dfrac{{20 - 42}}{5}}&{\dfrac{{ - 30}}{5}} \end{array}} \right] \\ 2X = \left[ {\begin{array}{*{20}{c}} {\dfrac{4}{5}}&{\dfrac{{ - 24}}{5}} \\ {\dfrac{{ - 22}}{5}}&{\dfrac{{ - 30}}{5}} \end{array}} \right] \\$
Now dividing by 2 we get:-
$X = \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{ - 12}}{5}} \\ {\dfrac{{ - 11}}{5}}&{\dfrac{{ - 15}}{5}} \end{array}} \right] \\ \Rightarrow X = \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{ - 12}}{5}} \\ {\dfrac{{ - 11}}{5}}&{ - 3} \end{array}} \right] \\$
Hence,
$X = \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{ - 12}}{5}} \\ {\dfrac{{ - 11}}{5}}&{ - 3} \end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{5}}&{\dfrac{{13}}{5}} \\ {\dfrac{{14}}{5}}&{\dfrac{{ - 10}}{5}} \end{array}} \right]$

Note:Students should note that when a matrix is multiplied by a scalar then every term inside the matrix gets multiplied by that scalar.Also, students should carefully calculate the values and use the elimination method to solve the equations.