Question

An amount of ₹65000 is invested in three investment at rate of 6%, 8% and 9% per annum respectively. The total annual income is₹4800.the income from the third investment is₹600 more than the income from the second investment. Using matrix algebra, determine the amount of each investment.
1

Answer 1

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 |

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

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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