Answer
351.9k+ views
Hint: To solve this question you should know about:
Linear equation of two variables: It is an equation of first order. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. When there are two variables then this equation will be a linear equation of two variables.
For example: $ax + by = c$
Matrix: It is a collection of numbers in a row and column format.
\[\left[ {\begin{array}{*{20}{c}}
2&4 \\
3&6
\end{array}} \right]\] is a matrix having $2 \times 2$ .
Complete step-by-step answer:
As given in question.
We have, $x + 2y = 4$ and $2x - y = 3$
Here we have to find the value of $x$ and $y$ .
Try it write in matrix form.
$AX = B$
$x + 2y = 4$
And
$2x - y = 3$
We can write it as,
\[\left[ {\begin{array}{*{20}{c}}
1&2 \\
2&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
4 \\
3
\end{array}} \right]\]
To solve we will change this equation into new one,
$X = {A^{ - 1}}B$
${A^{ - 1}}$ is inverse matrix of $A$
We calculate the inverse of $A$ .
${A^{ - 1}} = \dfrac{1}{{\det A}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
{ - 2}&1
\end{array}} \right]$
$\det A = 1 \times \left( { - 1} \right) - 2 \times 2$
$\det A = \left( { - 1} \right) - 4 = - 5$
Keeping it in above equation. We get,
\[{A^{ - 1}} = \dfrac{1}{{ - 5}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
{ - 2}&1
\end{array}} \right]\]
\[ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
{\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\]
Keeping it back in the equation;
$X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
{\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
4 \\
3
\end{array}} \right]$
By doing multiplication. We get,
$X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{5} \times 4 + \,\dfrac{2}{5} \times 3} \\
{\dfrac{2}{5} \times 4 + \,\dfrac{{ - 1}}{5} \times 3}
\end{array}} \right]$
$ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{4}{5} + \,\dfrac{6}{5}} \\
{\dfrac{8}{5} - \,\dfrac{3}{5}}
\end{array}} \right]$
\[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{10}}{5}} \\
{\dfrac{5}{5}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2 \\
1
\end{array}} \right]\]
By keeping value $X$ . We get,
$\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2 \\
1
\end{array}} \right]$
By equating. We will get,
$x = 2$ and $y = 1$ .
So, the correct answer is “ $ x = 2 $ and $ y = 1 $ ”.
Note: check the value $x = 2$ and $y = 1$ is true or not.
Keep in,
$
x + 2y = 4 \\
\Rightarrow 2 + 2(1) = 4 \;
$
Hence, for this is true.
Keep in,
$
2x - y = 3 \\
\Rightarrow 2(2) - 1 = 3 \\
\Rightarrow 4 - 1 = 3 \;
$
Hence, this value is true for this equation.
Linear equation of two variables: It is an equation of first order. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. When there are two variables then this equation will be a linear equation of two variables.
For example: $ax + by = c$
Matrix: It is a collection of numbers in a row and column format.
\[\left[ {\begin{array}{*{20}{c}}
2&4 \\
3&6
\end{array}} \right]\] is a matrix having $2 \times 2$ .
Complete step-by-step answer:
As given in question.
We have, $x + 2y = 4$ and $2x - y = 3$
Here we have to find the value of $x$ and $y$ .
Try it write in matrix form.
$AX = B$
$x + 2y = 4$
And
$2x - y = 3$
We can write it as,
\[\left[ {\begin{array}{*{20}{c}}
1&2 \\
2&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
4 \\
3
\end{array}} \right]\]
To solve we will change this equation into new one,
$X = {A^{ - 1}}B$
${A^{ - 1}}$ is inverse matrix of $A$
We calculate the inverse of $A$ .
${A^{ - 1}} = \dfrac{1}{{\det A}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
{ - 2}&1
\end{array}} \right]$
$\det A = 1 \times \left( { - 1} \right) - 2 \times 2$
$\det A = \left( { - 1} \right) - 4 = - 5$
Keeping it in above equation. We get,
\[{A^{ - 1}} = \dfrac{1}{{ - 5}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
{ - 2}&1
\end{array}} \right]\]
\[ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
{\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\]
Keeping it back in the equation;
$X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{ - 1}}{{ - 5}}}&{\dfrac{{ - 2}}{{ - 5}}} \\
{\dfrac{{ - 2}}{{ - 5}}}&{\dfrac{1}{{ - 5}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
4 \\
3
\end{array}} \right]$
By doing multiplication. We get,
$X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{5} \times 4 + \,\dfrac{2}{5} \times 3} \\
{\dfrac{2}{5} \times 4 + \,\dfrac{{ - 1}}{5} \times 3}
\end{array}} \right]$
$ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{4}{5} + \,\dfrac{6}{5}} \\
{\dfrac{8}{5} - \,\dfrac{3}{5}}
\end{array}} \right]$
\[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}
{\dfrac{{10}}{5}} \\
{\dfrac{5}{5}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2 \\
1
\end{array}} \right]\]
By keeping value $X$ . We get,
$\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2 \\
1
\end{array}} \right]$
By equating. We will get,
$x = 2$ and $y = 1$ .
So, the correct answer is “ $ x = 2 $ and $ y = 1 $ ”.
Note: check the value $x = 2$ and $y = 1$ is true or not.
Keep in,
$
x + 2y = 4 \\
\Rightarrow 2 + 2(1) = 4 \;
$
Hence, for this is true.
Keep in,
$
2x - y = 3 \\
\Rightarrow 2(2) - 1 = 3 \\
\Rightarrow 4 - 1 = 3 \;
$
Hence, this value is true for this equation.
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