Question

Solve the following equations for \[{\text{X}}\] and \[{\text{Y}}\], if \[3{\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\] , and \[{\text{X - 3Y = }}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\] .

Answer 1

Hint: Here we will solve this question by comparing both the equations for \[{\text{X}}\] and \[{\text{Y}}\].
Also, we will use the matrix property of subtraction which states that if we need to subtract two matrices as given below:
\[{\text{X = }}\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\] and \[{\text{Y = }}\left[ {\begin{array}{*{20}{c}}
  e&f \\
  g&h
\end{array}} \right]\] , then \[{\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  {a - e}&{b - f} \\
  {c - g}&{d - h}
\end{array}} \right]\].

Complete step by step solution:
Step 1: We have given the two equations as shown below:
\[3{\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\] …………………… (i)
\[{\text{X - 3Y = }}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\] ………………………. (ii)
By multiplying the equation (i) with \[3\], we get:
\[3\left( {3{\text{X - Y}}} \right){\text{ = 3}}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\]
By doing the multiplication into the LHS side, we get:
\[{\text{9X - 3Y = 3}}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\] ………………. (iii)
Step 2: By subtracting equation (iii) from equation (ii), we get:
\[\begin{gathered}
  {\text{9X - 3Y = 3}}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right] \\
  {\text{X - 3Y = }}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right] \\
  \overline {{\text{8X - 0 = 3}}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]} \\
\end{gathered} \]
By multiplying \[3\] inside the matrix \[\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right]\] into the RHS side, we get:
\[ \Rightarrow {\text{8X = }}\left[ {\begin{array}{*{20}{c}}
  3&{ - 3} \\
  { - 3}&3
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\]
By subtracting the matrices into the RHS side of the equation, we get:
\[ \Rightarrow {\text{8X = }}\left[ {\begin{array}{*{20}{c}}
  {3 - 0}&{ - 3 - \left( { - 1} \right)} \\
  { - 3 - 0}&{3 - \left( { - 1} \right)}
\end{array}} \right]\]
By opening the brackets inside the matrix and adding the elements in the equation \[{\text{8X = }}\left[ {\begin{array}{*{20}{c}}
  {3 - 0}&{ - 3 - \left( { - 1} \right)} \\
  { - 3 - 0}&{3 - \left( { - 1} \right)}
\end{array}} \right]\] , we get:
\[ \Rightarrow {\text{8X = }}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  { - 3}&4
\end{array}} \right]\]
By bringing \[8\] in to the RHS side of the equation for finding the value of \[{\text{X}}\] , we get:
\[ \Rightarrow {\text{X = }}\dfrac{1}{8}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  { - 3}&4
\end{array}} \right]\]
Finally, dividing \[8\] with every element inside the matrix, we get:
\[ \Rightarrow {\text{X = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{3}{8}}&{\dfrac{{ - 2}}{8}} \\
  {\dfrac{{ - 3}}{8}}&{\dfrac{4}{8}}
\end{array}} \right]\]
By doing the final division inside the matrix, we get:
\[ \Rightarrow {\text{X = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{3}{8}}&{\dfrac{{ - 1}}{4}} \\
  {\dfrac{{ - 3}}{8}}&{\dfrac{1}{2}}
\end{array}} \right]\]
Step 3: Similarly, for finding the value of \[{\text{Y}}\] we will again repeat the step number (1) and (2) as shown below:
By multiplying the equation (ii) with \[3\], we get:
\[3\left( {{\text{X - 3Y}}} \right){\text{ = 3}}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\]
By doing the multiplication into the LHS side, we get:
\[3{\text{X}} - 9{\text{Y = 3}}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\] ………………. (iv)
Step 4: By subtracting the equation (iv) from equation (i), we get:
\[\begin{gathered}
  3{\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right] \\
  {\text{3X - 9Y = 3}}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right] \\
  \overline {{\text{0 + 8Y = }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]} \\
\end{gathered} \]
By multiplying \[3\] inside the matrix \[\left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  0&{ - 1}
\end{array}} \right]\] into the RHS side, we get:
\[ \Rightarrow {\text{8Y = }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 1}&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  0&{ - 3} \\
  0&{ - 3}
\end{array}} \right]\]
By subtracting the matrices into the RHS side of the equation, we get:
\[ \Rightarrow {\text{8Y = }}\left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{ - 1 - \left( { - 3} \right)} \\
  { - 1 - 0}&{1 - \left( { - 3} \right)}
\end{array}} \right]\]
By opening the brackets inside the matrix and adding the elements in the equation \[{\text{8Y = }}\left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{ - 1 - \left( { - 3} \right)} \\
  { - 1 - 0}&{1 - \left( { - 3} \right)}
\end{array}} \right]\] , we get:
\[ \Rightarrow {\text{8Y = }}\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&4
\end{array}} \right]\]
By bringing \[8\] in to the RHS side of the equation for finding the value of \[{\text{Y}}\] , we get:
\[ \Rightarrow {\text{Y = }}\dfrac{1}{8}\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&4
\end{array}} \right]\]
Finally, dividing \[8\] with every element inside the matrix, we get:
\[ \Rightarrow {\text{Y = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{8}}&{\dfrac{2}{8}} \\
  {\dfrac{{ - 1}}{8}}&{\dfrac{4}{8}}
\end{array}} \right]\]
By doing the final division inside the matrix, we get:
\[ \Rightarrow {\text{Y = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{8}}&{\dfrac{1}{4}} \\
  {\dfrac{{ - 1}}{8}}&{\dfrac{1}{2}}
\end{array}} \right]\]

Therefore \[\because \] \[{\text{Y = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{8}}&{\dfrac{1}{4}} \\
  {\dfrac{{ - 1}}{8}}&{\dfrac{1}{2}}
\end{array}} \right]\] and \[{\text{X = }}\left[ {\begin{array}{*{20}{c}}
  {\dfrac{3}{8}}&{\dfrac{{ - 1}}{4}} \\
  {\dfrac{{ - 3}}{8}}&{\dfrac{1}{2}}
\end{array}} \right]\]

Note: Students need to take care while solving the addition, subtraction or product of any two matrices. While doing the multiplication you should remember the below points:
The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.
And the result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix.