Courses
Courses for Kids
Free study material
Free LIVE classes
More
LIVE
Join Vedantu’s FREE Mastercalss

If ${\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&0 \\
  1&0&0
\end{array}} \right]$and${\text{X + Y = }}\left[ {\begin{array}{*{20}{c}}
  3&5&1 \\
  { - 1}&1&4 \\
  {11}&8&0
\end{array}} \right]$, find the value of X and Y.

Answer
VerifiedVerified
364.8k+ views
Hint- Use matrix addition/subtraction and multiplication with scalar while simplifying RHS and perform elimination methods.

Two linear equations are given to us that is
\[{\text{X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&0 \\
  1&0&0
\end{array}} \right]\]……………………………… (1)
\[{\text{X + Y = }}\left[ {\begin{array}{*{20}{c}}
  3&5&1 \\
  { - 1}&1&4 \\
  {11}&8&0
\end{array}} \right]\]……………………….. (2)
We will be simply using the elimination method to solve for the value of X and Y but in the right hand side we have been provided with a \[{\text{3}} \times {\text{3}}\]matrix, so in RHS we will be using matrix addition.
Matrix addition states that if we have two \[{\text{m}} \times {\text{n}}\]matrix like \[\left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\]and \[\left[ {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\
  {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \\
  {{b_{31}}}&{{b_{32}}}&{{b_{33}}}
\end{array}} \right]\]then
The addition of these two matrix will be \[\left[ {\begin{array}{*{20}{c}}
  {{a_{11}} + {b_{11}}}&{{a_{12}} + {b_{12}}}&{{a_{13}} + {b_{13}}} \\
  {{a_{21}} + {b_{21}}}&{{a_{22}} + {b_{22}}}&{{a_{23}} + {b_{23}}} \\
  {{a_{31}} + {b_{31}}}&{{a_{32}} + {b_{32}}}&{{a_{33}} + {b_{33}}}
\end{array}} \right]\]
Hence now let’s add equation 1 and equation 2 so we get
\[{\text{X - Y}} + {\text{X + Y}} = \left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&0 \\
  1&0&0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  3&5&1 \\
  { - 1}&1&4 \\
  {11}&8&0
\end{array}} \right]\]
\[2X{\text{ = }}\left[ {\begin{array}{*{20}{c}}
  4&6&2 \\
  0&2&4 \\
  {12}&8&0
\end{array}} \right]\]
Now \[{\text{X = }}\dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  4&6&2 \\
  0&2&4 \\
  {12}&8&0
\end{array}} \right]\]or \[{\text{X = }}\left[ {\begin{array}{*{20}{c}}
  2&3&1 \\
  0&1&2 \\
  6&4&0
\end{array}} \right]\]
Now let’s subtract equation (2) from equation (1)
\[{\text{X - Y - X - Y = }}\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&0 \\
  1&0&0
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  3&5&1 \\
  { - 1}&1&4 \\
  {11}&8&0
\end{array}} \right]\]
Let’s simplify this further we get
\[ - 2Y = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 4}&0 \\
  2&0&{ - 4} \\
  { - 10}&{ - 8}&0
\end{array}} \right]{\text{ or Y = }}\dfrac{{ - 1}}{2}\left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 4}&0 \\
  2&0&{ - 4} \\
  { - 10}&{ - 8}&0
\end{array}} \right]\]
Thus \[Y = \left[ {\begin{array}{*{20}{c}}
  1&2&0 \\
  { - 1}&0&2 \\
  5&4&0
\end{array}} \right]\]

Note-Such problems could be solved via the concept that we use while solving any two linear equations. We could have even used the method of substitution instead of elimination to solve this, the only thing which needs to be taken care of is matrix addition and subtraction is a bit different from simpler linear addition and subtraction.

Last updated date: 26th Sep 2023
Total views: 364.8k
Views today: 4.64k