# Solve \[2x - y = 7;{\text{ }}3x - 2y = 11\] by using the matrix inversion method.

Answer

Verified

384k+ 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

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many meters are there in a kilometer And how many class 8 maths CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

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