Answer to Question #9561 in Math for Candice

Question

Sub 25
Ques. 5

Solve this system:

1/r + 3/s - 2/t = 1

2/r + 3/s - 4/t = 1

1/r - 6/s - 6/t = 0

Answer 1

Make a substitution
x = 1/r
y = 1/s
z = 1/t

Then we will
have the folowing system:

x + 3y - 2z = 1
2x + 3y - 4z =
1
x - 6y - 6z = 0

Let us use Jordan method.
The matrix of the
system is

1 3 -2 1
2 3 -4 1
1 -6 -6 0

Make a
transformation, where R[i] is the i-th row:

R[2] = R[2] - 2*R[1]
R[3]
= R[3] - R[1]

1 3 -2 1
0 -3 0 -1
0 -9 -4 -1

R[1]
= R[1]+R[2]
R[3] = R[3]-3*R[2]


1 0 -2 0
0 -3 0
-1
0 0 -4 2


R[2]=R[2]/(-3)
R[3]=R[3]/(-4)


1
0 -2 0
0 1 0 1/3
0 0 1 -1/2

R[1] =
R[1]+2*R[3]

1 0 0 -1
0 1 0 1/3
0 0 1
-1/2


Thus

x = -1, y = 1/3, z = -1/2

Hence

r =
1/x = -1
s = 1/y = 3
t = 1/z = -2