Keirnath S. answered • 03/29/21
Effective and Specifically Skilled in chemistry, math, and engineering
F= francs ; S= shillings ; M= marks
First set up the three known equations with the three unknown variables F, S, M
equation 1: we know he has a total of 33 coins therefore
F + S + M = 33
equation 2: we know the value of each coin and the total amount they add up to is $6.33 therefore
.02F + .14S + .25M = 6.33
equation 3: we also know that it take 3 times as many shilling to equal as many marks as he has
0F + 3S -1M = 0 to collaborate this, manipulate the equation to check for accuracy 3S=1M or S=1/3M
Now that we have all equations set up we can utilize the coefficients in matrix A to solve the linear algebra problem with Cramer's Rule where Ax=b in that x= A^-1b and X1=F, X2=S, and X3=M
b
| 1 1 1 | |x1| = | 33 |
A= |.02 .14 .25 | |x2| = |6.33|
| 0 3 -1 | |x2| = |0.00|
Now Find A1, A2, A3 by replacing the each column in the matrix with column b as seen below
b
| 33 1 1 |
A1= |6.33 .14 .25 |
| 0 3 -1 |
b
| 1 33 1 |
A2= |.02 6.33 .25 |
| 0 0 -1 |
b
| 1 1 33 |
A3= |.02 .14 6.33 |
| 0 3 0 |
Now that we found A1 A2 and A3 we use Cramer's rule formula
x1=det(A1)/det(A)
x2=det(A2)/det(A)
x3=det(A3)/det(A)
det(3x3)=a * det |e f| - b* det |d f| + c* det |d e|
|h i| |g i| |g h|
where:
|a b c|
|d e f |
|g h i |
Example for solving X3
detA3= 1*(.14*0 - 3*6.33) - 1*(.02*0 - 6.33*0) + 33*(.02*3 - .14*0) = -17.01
det A = 1*(.14*-1 -.25*3) -1*(.02*-1 -.25*0) + 1*(.02*3 -.14*0) = -.81
x3=detA3/detA=-17.01/-.81= 21
from there it is most efficient to solve equation 3, rather than solving for all the other determinants. Therefore,
0*F +3*S + -1*21=0. And S=7
Or there is 3 times more Marks than shillings so 21/3=7
Then use x3 & x2 to solve for x1 using equation 1
so 33- 21- 7= 5 francs
now all of our equations are coherent
Answer:
X1=Francs=5
X2=Shillings=7
X3=Marks=21
