Question

Find x and y , if
$
  \left( i \right)x + y = \left( {\begin{array}{*{20}{c}}
  7&0 \\
  2&5
\end{array}} \right),x - y = \left( {\begin{array}{*{20}{c}}
  3&0 \\
  0&3
\end{array}} \right) \\
  \left( {ii} \right)2x + 3y = \left( {\begin{array}{*{20}{c}}
  2&3 \\
  4&0
\end{array}} \right),3x + 2y = \left( {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 1}&5
\end{array}} \right) \\
$

Answer 1

Hint: Here we have given equations in x and y and we have to find x and y on solving these equations. here we have to care about addition of matrices that the first element of the first row and first column of the first matrix must be added with the first element of first row and first column of second matrix.

Complete step-by-step answer:
We have given
$\left( i \right)x + y = \left( {\begin{array}{*{20}{c}}
  7&0 \\
  2&5
\end{array}} \right),x - y = \left( {\begin{array}{*{20}{c}}
  3&0 \\
  0&3
\end{array}} \right)$
When we add both the equations we can easily get the value of x and then on putting the value of x we get the value of y and here just we have to care about addition of these matrices.
On adding we get
2x= $\left( {\begin{array}{*{20}{c}}
  {10}&0 \\
  2&8
\end{array}} \right)$
And hence
X= $\left( {\begin{array}{*{20}{c}}
  5&0 \\
  1&4
\end{array}} \right)$
And for y we put the value of x then we can get y
And hence
Y = $\left( {\begin{array}{*{20}{c}}
  2&0 \\
  1&1
\end{array}} \right)$
And now second question we have
$\left( {ii} \right)2x + 3y = \left( {\begin{array}{*{20}{c}}
  2&3 \\
  4&0
\end{array}} \right),3x + 2y = \left( {\begin{array}{*{20}{c}}
  2&{ - 2} \\
  { - 1}&5
\end{array}} \right)$
Here on solving these equations we have to find x and y.
On multiplying by 3 on equation 1 we get
$6x + 9y = \left( {\begin{array}{*{20}{c}}
  6&9 \\
  {12}&0
\end{array}} \right)$ eq.1
On multiplying by 2 in equation 2 we get,
$6x + 4y = \left( {\begin{array}{*{20}{c}}
  4&{ - 4} \\
  { - 2}&{10}
\end{array}} \right)$ eq.2
On subtracting eq.2 from eq.1 we get,
5y=$\left( {\begin{array}{*{20}{c}}
  2&{13} \\
  {14}&{ - 10}
\end{array}} \right)$
And hence
Y= $\left( {\begin{array}{*{20}{c}}
  {\dfrac{2}{5}}&{\dfrac{{13}}{5}} \\
  {\dfrac{{14}}{5}}&{ - 2}
\end{array}} \right)$
Now on putting the value of y we can get easily x.
And hence x = $\left( {\begin{array}{*{20}{c}}
  {\dfrac{2}{5}}&{\dfrac{{ - 12}}{5}} \\
  {\dfrac{{ - 11}}{5}}&3
\end{array}} \right)$

Note: Whenever we get this type of question the key concept of solving is we have to have knowledge of algebra of matrices. and we should understand that these questions are the same as linear equations in two variables just one thing changes that is algebra of matrices.