
Solve \[2x - y = 7;{\text{ }}3x - 2y = 11\] by using the matrix inversion method.
Answer
560.1k+ views
Hint:- Write equations in form of \[AX = B\]. Here A is a square matrix and its inverse is $A^{-1}$. Matrix inversion method is applied to non-singular square matrix.
As given in the question to solve the given equations using matrix inversion method,
When there is said to solve using matrix inversion method then we had to,
First of all write the system of equations in the form of \[AX = B\].
Where, A will be a matrix containing coefficients of variables of a given equation.
Where, B will be a matrix containing constant terms of the given equations.
And X will be a matrix containing variables of the given equations.
Let the equations will be,
\[ \Rightarrow ax + by = c\] and \[dx + ey = f\]
Then, \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
d&e
\end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
c \\
f
\end{array}} \right]\]
So, if the given equations be.
\[ \Rightarrow 2x - y = 7\] (1)
\[ \Rightarrow 3x - 2y = 11\] (2)
So, solving equation 1 and 2 using matrix inversion method. We get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right]\] (i.e.) \[AX = B\]
\[ \Rightarrow X = {A^{ - 1}}B\] (3)
Where \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
d&e
\end{array}} \right];X = \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right]\]
Now, we had to find \[{A^{ - 1}}\].
As, we know that \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)\].
Where \[\left| A \right|\] is the determinant of \[A\] and,
\[ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right| = 2*( - 2) - (3)*( - 1) = - 1\]
And as we know that for any matrix, \[C = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]\].
\[ \Rightarrow adj(C) = \left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right]\]
\[ \Rightarrow \]So, \[adj(A) = \left[ {\begin{array}{*{20}{c}}
{ - 2}&1 \\
{ - 3}&2
\end{array}} \right]\]
\[ \Rightarrow \]Hence, \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A) = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
{ - 2}&1 \\
{ - 3}&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\]
Now, putting value of \[{A^{ - 1}}\] and \[B\] in the equation 3 we get,
\[
\Rightarrow X = {A^{ - 1}}B = \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{14 - 11} \\
{21 - 22}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 1}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 1}
\end{array}} \right] \\
\]
So, on comparing we get \[x = 3\] and \[y = - 1\].
Note:- Whenever we came up with this type of problem then, first write the given
Linear equations in form of \[AX = B\], And then find the value of \[{A^{ - 1}}\] by using formula
\[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)\] and then multiply \[{A^{ - 1}}\] by \[B\]. Then you will get required value of the Matrix \[X\], which gives the value of all variables.
As given in the question to solve the given equations using matrix inversion method,
When there is said to solve using matrix inversion method then we had to,
First of all write the system of equations in the form of \[AX = B\].
Where, A will be a matrix containing coefficients of variables of a given equation.
Where, B will be a matrix containing constant terms of the given equations.
And X will be a matrix containing variables of the given equations.
Let the equations will be,
\[ \Rightarrow ax + by = c\] and \[dx + ey = f\]
Then, \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
d&e
\end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
c \\
f
\end{array}} \right]\]
So, if the given equations be.
\[ \Rightarrow 2x - y = 7\] (1)
\[ \Rightarrow 3x - 2y = 11\] (2)
So, solving equation 1 and 2 using matrix inversion method. We get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right]\] (i.e.) \[AX = B\]
\[ \Rightarrow X = {A^{ - 1}}B\] (3)
Where \[A = \left[ {\begin{array}{*{20}{c}}
a&b \\
d&e
\end{array}} \right];X = \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right]\]
Now, we had to find \[{A^{ - 1}}\].
As, we know that \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)\].
Where \[\left| A \right|\] is the determinant of \[A\] and,
\[ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right| = 2*( - 2) - (3)*( - 1) = - 1\]
And as we know that for any matrix, \[C = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right]\].
\[ \Rightarrow adj(C) = \left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right]\]
\[ \Rightarrow \]So, \[adj(A) = \left[ {\begin{array}{*{20}{c}}
{ - 2}&1 \\
{ - 3}&2
\end{array}} \right]\]
\[ \Rightarrow \]Hence, \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A) = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
{ - 2}&1 \\
{ - 3}&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\]
Now, putting value of \[{A^{ - 1}}\] and \[B\] in the equation 3 we get,
\[
\Rightarrow X = {A^{ - 1}}B = \left[ {\begin{array}{*{20}{c}}
2&{ - 1} \\
3&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
7 \\
{11}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{14 - 11} \\
{21 - 22}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 1}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3 \\
{ - 1}
\end{array}} \right] \\
\]
So, on comparing we get \[x = 3\] and \[y = - 1\].
Note:- Whenever we came up with this type of problem then, first write the given
Linear equations in form of \[AX = B\], And then find the value of \[{A^{ - 1}}\] by using formula
\[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)\] and then multiply \[{A^{ - 1}}\] by \[B\]. Then you will get required value of the Matrix \[X\], which gives the value of all variables.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Pomato is a Somatic hybrid b Allopolyploid c Natural class 12 biology CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

An example of ex situ conservation is a Sacred grove class 12 biology CBSE

What is the chemical name of rust Write its formul class 12 chemistry CBSE
