Question

An amount of Rs. 65,000 is invested in three investments at the rate of 6%, 8% and 9% per annum, respectively. The total annual income is Rs. 4,800. The income from the third investment is Rs. 600 more than the income from second investment. Using matrix algebra, determine the amount of each investment

Answer 1

Solution using Matrix Algebra by Cramer's Rule..

Answer 2

Given :  An amount of Rs. 65,000 is invested in three investments at the rate of 6%, 8% and 9% per annum, respectively. The total annual income is Rs. 4,800. The income from the third investment is Rs. 600 more than the income from second investment.

To find :   Using matrix algebra, determine the amount of each investment

Solution:

Let say amount of investment  at  6%, 8% and 9% per annum,

is x , y & z respectively

x + y + z  = 65000

6x/100 + 8y/100 + 9z/100 = 4800

=> 6x + 8y + 9z  =  480000

8y/100  + 600 = 9z/100

=> 8y  -  9z  =  - 60000

x +    y      + z  =   65000

6x +  8y   + 9z  =  480000

0*x + 8y -  9z  =  - 60000

$\left[\begin{array}{ccc}1&1&1\\6&8&9\\0&8&-9\end{array}\right]$  $\left[\begin{array}{c} x\\y\\z\end{array}\right]$    $= \left[\begin{array}{c} 65000\\480000\\-60000\end{array}\right]$

$A = \left[\begin{array}{ccc}1&1&1\\6&8&9\\0&8&-9\end{array}\right]$

 $A \left[\begin{array}{c} x\\y\\z\end{array}\right]$    $= \left[\begin{array}{c} 65000\\480000\\-60000\end{array}\right]$

=>  $\left[\begin{array}{c} x\\y\\z\end{array}\right]$   $= A^{-1} \left[\begin{array}{c} 65000\\480000\\-60000\end{array}\right]$

| A |  =  1 ( 8 *( -9 )- 8 * 9 )  - 1( 6 * (-9) - 0 ) + 1 ( 6 * 8 - 0 )

= -144 + 54 + 48

= -42

A⁻¹  =  AdjA / | A |

$Adj A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$

$A = \left[\begin{array}{ccc}1&1&1\\6&8&9\\0&8&-9\end{array}\right]$

A₁₁  = (-1)¹⁺¹ (8*(-9) - 8*(9)) = -144   

A₁₂ =  (-1)¹⁺² (6*(-9) - 0*(9)) = 54

A₁₃ = (-1)¹⁺³(6*8 - 0*(8)) =  48

A₂₁  = (-1)²⁺¹ (1*(-9) - 8*(1)) = 17

A₂₂ =  (-1)²⁺² (1*(-9) - 0*(1)) = -9

A₂₃ = (-1)²⁺³(1*8 - 0*(1)) = -8

A₃₁  = (-1)³⁺¹ (1*9 - 8*(1)) = 1

A₃₂ =  (-1)³⁺² (1*(9) - 6*(1)) = -3

A₃₃ = (-1)³⁺³(1*8 - 6*(1)) = 2

$A^{-1} = \frac{-1}{42} \begin{bmatrix} -144 &17 & 1 \\ 54 & -9 & -3 \\ 48 & -8 & 2 \end{bmatrix}$

$\left[\begin{array}{c} x\\y\\z\end{array}\right]$   $= \frac{-1}{42} \begin{bmatrix} -144 &17 & 1 \\ 54 & -9 & -3 \\ 48 & -8 & 2 \end{bmatrix}$    $\left[\begin{array}{c} 65000\\480000\\-60000\end{array}\right]$  

x = (-1/42) ( -144 * 65000 + 17*480000 + 1 *(-60000))  = 30000

y =   (-1/42) (54 * 65000 - 9*480000- 3  *(-60000) )  = 15000

z =   (-1/42) (48 * 65000 - 8*480000 + 2  *(-60000) )  = 20000

Incomes are 30000 , 15000 , 20000  

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