Question

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 1

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.