# Answer to Question #9561 in Other Math for Candice

Question #9561

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

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

Expert's answer

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

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

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