Question

Find the matrices $X$ and $Y$ , if $2X - Y = \left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right]$ and $X + 2Y = \left[ {\begin{array}{*{20}{c}}
  3&2&5 \\
  { - 2}&1&{ - 7}
\end{array}} \right]$

Answer 1

Hint: This problem can be treated as a linear equation in two variables in form of $X$ and $Y$ and solved just like how we solve a linear equation


Complete step-by-step answer:
The given matrices are
$2X - Y = \left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right] \cdots \left( 1 \right)$
$X + 2Y = \left[ {\begin{array}{*{20}{c}}
  3&2&5 \\
  { - 2}&1&{ - 7}
\end{array}} \right] \cdots \left( 2 \right)$
For performing the addition or subtraction of two matrices, the order of the matrices should be the same . The order of both the matrices is $2 \times 3$ .
Let’s suppose the matrices be $X = \left[ {\begin{array}{*{20}{c}}
  p&q&r \\
  s&t&u
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f
\end{array}} \right]$
After analyzing the two equations, it is clear that matrix $Y$ can be eliminated for the calculation of matrix $X$ by multiplying equation (1) by 2 and adding equation (1) and (2).
Multiplying equation (1) by (2),
$2\left( {2X - Y} \right) = 2\left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right]$
If a matrix is multiplied by a scalar quantity, then the scalar is to be multiplied by each and every term of the matrix as shown below,
$4X - 2Y = \left[ {\begin{array}{*{20}{c}}
  {12}&{ - 12}&0 \\
  { - 8}&4&2
\end{array}} \right] \cdots \left( 3 \right)$
Adding equation (2 ) and (3)
\[\left( {X + 2Y} \right) + \left( {4X - 2Y} \right) = \left[ {\begin{array}{*{20}{c}}
  3&2&5 \\
  { - 2}&1&{ - 7}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {12}&{ - 12}&0 \\
  { - 8}&4&2
\end{array}} \right]\]
For adding the two matrices , add their corresponding terms.
$\begin{gathered}
  5X = \left[ {\begin{array}{*{20}{c}}
  {3 + 12}&{2 - 12}&{5 - 0} \\
  { - 2 - 8}&{1 + 4}&{ - 7 + 2}
\end{array}} \right] \\
  5X = \left[ {\begin{array}{*{20}{c}}
  {15}&{ - 10}&5 \\
  { - 10}&5&{ - 5}
\end{array}} \right] \cdots \left( 4 \right) \\
\end{gathered} $
Now it is clear that in equation (4), we can take out 5 as a common factor.
$5X = 5\left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&1 \\
  { - 2}&1&{ - 1}
\end{array}} \right] \cdots \left( 5 \right)$
Cancelling 5 from both sides of LHS and RHS in equation (5), matrix $X$ is obtained as
$X = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&1 \\
  { - 2}&1&{ - 1}
\end{array}} \right] \cdots \left( 6 \right)$
Substitute the value of matrix $X$ in equation (1),
\[2\left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&1 \\
  { - 2}&1&{ - 1}
\end{array}} \right] - Y = \left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right] \cdots \left( 7 \right)\]
Multiply all the terms of the matrix $X$ by 2 as given in the equation (7)
\[\left[ {\begin{array}{*{20}{c}}
  6&{ - 4}&2 \\
  { - 4}&2&{ - 2}
\end{array}} \right] - Y = \left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right]\]
Now rearrange the terms and calculate the value of matrix $Y$ as,
$Y = \left[ {\begin{array}{*{20}{c}}
  6&{ - 4}&2 \\
  { - 4}&2&{ - 2}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  6&{ - 6}&0 \\
  { - 4}&2&1
\end{array}} \right]$
Now subtract the 2 matrices by subtracting their corresponding terms,
$Y = \left[ {\begin{array}{*{20}{c}}
  0&2&2 \\
  0&0&{ - 3}
\end{array}} \right]$
Hence, the value of matrix $X = \left[ {\begin{array}{*{20}{c}}
  3&{ - 2}&1 \\
  { - 2}&1&{ - 1}
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  0&2&2 \\
  0&0&{ - 3}
\end{array}} \right]$..


Note: The important concepts to be remembered are
1)Two matrices can be added or subtracted only when they have the same order.
2)Multiplication of a matrix by a scalar, leads to multiplication of its terms.