# The monthly incomes of Aryan and Babban are in the ratio \[3:4\] and their monthly expenditures are in the ratio \[5:7\]. If each saves \[{\text{Rs}}.15000\]per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer

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Hint: Convert the rations into a system of equations of linear equations and then solve using Gaussian Elimination method.

Let the incomes of Aryan and Babban be $3x$ and $4x$ respectively.

In the same way, their expenditures would be $5y$ and $7y$ respectively.

Since Aryan and Babban saves \[{\text{Rs}}.15000\] , we get

$

3x - 5y = 15000.........................................\left( 1 \right) \\

4x - 7y = 15000........................................\left( 2 \right) \\

$

These are the systems of linear equations in the form $AX = B$, which can be solved by using Gaussian Elimination method.

So, $A = \left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

4&{ - 7}

\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]{\text{ and }}X = \left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right]$

This can be written in matrix form as

$\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

4&{ - 7}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_2} \to {R_2} - \dfrac{4}{3}{R_1}$

We get $\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

0&{ - \dfrac{1}{3}}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{ - 5000}

\end{array}} \right]$

Applying the row operation ${R_2} \to {R_2} \times - 3$

We get $\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_1} \to {R_1} + 5{R_2}$

We get $\left[ {\begin{array}{*{20}{c}}

3&0 \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{90000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_1} \to {R_1} \times \dfrac{1}{3}$

We get $\left[ {\begin{array}{*{20}{c}}

1&0 \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{30000} \\

{15000}

\end{array}} \right]$

Converting back into system of equations we have

$

x = 30000 \\

y = 15000 \\

$

Thus, the income of Aryan is $3x = 3 \times 30000 = 90000$ and the income of Babban is $4x = 4 \times 30000 = 120000$.

Hence incomes of Aryan and Babban are ${\text{Rs}}.90,000{\text{ and Rs}}.1,20,000$ respectively.

The value of the problem is Aryan is more interested in saving money than Babban. So, one must save money, no matter how much one earns.

Note: In this problem the matrix method is used to solve the system of $n$ linear equations in $n$ unknowns. Some types of matrix methods to solve the system of linear equations are Gaussian Elimination method, Inverse matrix method and Cramer's rule. Here we have used the Gaussian Elimination method.

Let the incomes of Aryan and Babban be $3x$ and $4x$ respectively.

In the same way, their expenditures would be $5y$ and $7y$ respectively.

Since Aryan and Babban saves \[{\text{Rs}}.15000\] , we get

$

3x - 5y = 15000.........................................\left( 1 \right) \\

4x - 7y = 15000........................................\left( 2 \right) \\

$

These are the systems of linear equations in the form $AX = B$, which can be solved by using Gaussian Elimination method.

So, $A = \left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

4&{ - 7}

\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]{\text{ and }}X = \left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right]$

This can be written in matrix form as

$\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

4&{ - 7}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_2} \to {R_2} - \dfrac{4}{3}{R_1}$

We get $\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

0&{ - \dfrac{1}{3}}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{ - 5000}

\end{array}} \right]$

Applying the row operation ${R_2} \to {R_2} \times - 3$

We get $\left[ {\begin{array}{*{20}{c}}

3&{ - 5} \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{15000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_1} \to {R_1} + 5{R_2}$

We get $\left[ {\begin{array}{*{20}{c}}

3&0 \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{90000} \\

{15000}

\end{array}} \right]$

Applying the row operation ${R_1} \to {R_1} \times \dfrac{1}{3}$

We get $\left[ {\begin{array}{*{20}{c}}

1&0 \\

0&1

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

x \\

y

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{30000} \\

{15000}

\end{array}} \right]$

Converting back into system of equations we have

$

x = 30000 \\

y = 15000 \\

$

Thus, the income of Aryan is $3x = 3 \times 30000 = 90000$ and the income of Babban is $4x = 4 \times 30000 = 120000$.

Hence incomes of Aryan and Babban are ${\text{Rs}}.90,000{\text{ and Rs}}.1,20,000$ respectively.

The value of the problem is Aryan is more interested in saving money than Babban. So, one must save money, no matter how much one earns.

Note: In this problem the matrix method is used to solve the system of $n$ linear equations in $n$ unknowns. Some types of matrix methods to solve the system of linear equations are Gaussian Elimination method, Inverse matrix method and Cramer's rule. Here we have used the Gaussian Elimination method.

Last updated date: 01st Oct 2023

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