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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.

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