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
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
•
Total views: 364.8k
•
Views today: 4.64k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
