SOLUTION: An amount of Rs. 65000 is invested in three investments at the rate of 6%, 8% and 9% per annum, repectiveliy. they total annual income is Rs. 4800. The income from the third invest

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Question 1161954: An amount of Rs. 65000 is invested in three investments at the rate of 6%, 8% and 9% per annum, repectiveliy. they total annual income is Rs. 4800. 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 by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x - 6% investment.
y = 8% investment.
z = 9% investment.

you have 3 equations to solve simultaneously.
they are:

x + y + z = 65000
.06*x + .08*y + .09*z = 4800
.09*z = .08*y + 600

that third equation is rearranged to become:
.08*y - .09*z = -600

your 3 equations are now:

x + y + z = 65000
.06*x + .08*y + .09*z = 4800
.08*y - .09*z = -600

put the coefficients of the variable terms and the constant terms into a row by 4 column matrix as shown below: 

                     1              1           1            65000
                     .06            .08         .09          4800
                     0              .08         -.09         -600


use the gauss elimination method to turn this matrix into the following: 

                    42              0            0         1,260,000
                    0               14           0         210,000
                    0               0           -21        -420,000


the first column is the coefficient of the x variable.
the second column is the coefficient of the y variable.
the third column is the coefficient of the z variable
the fourth column is the constant term.

solve for x in the first row to get x = 1,260,000 / 42 = 30,000
solve for y in the second row to get y = 210,000 / 14 = 15,000
solve for z in the third row to ge t z = -420,000 / -21 = 20,000

this was a headache to do manually, but i did it and got those results.

i also used an online gauss jordan calculator to do this mechanically and got the same results.

my manual effort was done in different steps than the mechanical effort, but the results were the same, as they should have been.

here are the results of the mechanized method.



here are the results of the manual method.




my manual effort did not go as far as the mechanical method because i didn't reduce the matrix to where all the coefficients were 1, but that didn't really matter since the same answer was gotten either way.

there are two basic methods.
gauss elimination and gauss jordan elimination.
gauss elimination is similar to what i did manually.
gauss jordan elimination goes one step further and makes all the coefficients equal to 1 so you can then just read off the answer.

my manual effort was pretty close to gauss jorden elimination with the exception that i didn't make the coefficients equal to 1.
at the point that i was at, i could have easily done that by doing the following operations.
R1 = R1 / 42
R2 = R2 / 14
R3 = R3 / -2121

the final matrix would have then been: 

                       1          0         0         30,000
                       0          1         0         15,000
                       0          0         1         20,000


here's the link to the calculator i used for the gauss-jordan elimination method.

http://www.gregthatcher.com/Mathematics/GaussJordan.aspxhttp://www.gregthatcher.com/Mathematics/GaussJordan.aspx

note that this calculator forced me to enter fractions rather than decimals, i.e. .08 become 8/100, etc.