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

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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Answer 2

Answer:

X= 30,000 Y = 15,000 Z= 20,000

HELLO DEAR,

Given : Total amount of invested in three investments at the rate of 6%, 8% ,9% per annum = Rs. 65,000.

And total annual income by above investment = Rs.4,800.

Solution: Let the three investment are X , Y and Z.

As per question.

X+ Y + Z = 65,000

and ,

(X× 6%) + (Y+8%) + ( Z× 9%)= 4,800

(X×6/100) + ( Y+8/100) + (Z× 9/100) = 4,800

( 3X/50) + ( 2Y/25) + (9Z/100) = 4,800

multiplying by 100 both side.

6X+ 8Y + 9Z= 480000 ................(2)

The third investment is Rs 600 more than the income from second investment can be written in form of mathetic,

(9Z/100)= (2Y/25) + 600

multiplying by 100.

9Z= 8Y + 60000

- 8Y + 9Z= 60000 ............(3)

Thier are three equation .

X+ Y + Z= 65,000 ..............(1)

6X+ 8Y + 9Z= 4,80,000...........(2)

0X+ 8Y -9 Z= -60,000...............(3).

By using matrix algebra we determine the value of X, Y and Z which is the amount of each investment.

Lets get start,

The above equation information can be written in the form of matrix as follow,

[ 1 1 1 ] [ X ] = [ 65000. ]

| 6 8 9 | | Y | = | 480000 |

| 0 8 -9| | Z |. = | -60000 |

where A = | 1 1 1 |

| 6 8 9|

| 0 8 -9|

| X | | 65000 |

| Y | = | 480000|

| Z | | -60000 |

| X | | 65000. |

| Y | = A^-1 | 480000 |

| Z | = | -60000 |

So, A^-1 is called inverse of a matrix.

and A^-1 is obtained by first finding the cofactor of A and then transpose of it.

( all remaining solution is done in above pic)

I HOPE IT HELP YOU DEAR,

THANKS.