Question

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

Answer 1

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.

Complete step-by-step answer:
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.