# 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

Verified

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.

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

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Total views: 364.8k

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Views today: 4.64k

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