Question

Solve The Equation By Cramer's Rule:

Answer 1

Thx for the points, have a nice dayyy

Answer 2

Answer:

Question:

x+y+z=3; x-y+z=1; x+y-2z=0

Step-by-step explanation:

-----» Given equation can be written as..

(1)x +(1)y+ (1)z =3

(1)x- (1)y+ (1)z =1

(1)x+ (1)y- (2)z =0

Step 1: We have To calculate D

$\: \: \: \: \: \: \: \: \: |1 \: \: \: \: 1 \: \: \: \: \: 1 \: \: | \\ d \: = |1 \: - 1 \: \: \: \: \: 1| \\ \: \: \: \: \: \: \: \: \: \: \: |1 \ \: \: 1 \: \: \: - 2|$

=1[(-1)(-2)-(1)(1)]-1[(1)(-2)-(1)(1)]+1[(1)(1)-(-1)(1)]

=1[2-1] -1[-2-1] +1[1+1]

=1(1) -1(-3) +1(2)

=1+3+2

D=6

______________________________________

Step 2: We have to calculate Dx!!

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: |3 \: \: \: \: \: 1 \: \: \: \: \: \: 1| \\ DX=|1\: \: - 1 \: \: \: 1 | \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: |0 \: \: 1 \: \: - 2|$

=3[(-1)(-2)-(1)(1)] -1[(1)(-2)-(1)(0)]+1[(1)(1)-(-1)(0)]

=3(2-1) -1(-2-0) +1(1-0)

=3(1) -1(-2) +1(1)

=3+2+1

DX=6

______________________________________

Step 3: We have to calculate Dy!

$\: \: \: \: \: \: \: \: \: \: \: |1 \: \: \: \: 3 \: \: \: \: \: \: 1| \\dy= |1 \: \: \: \: \: 1 \: \: \: \: \: 1| \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: |1 \: \: \: \: \: 0 \: \: - 2|$

=1[(1)(-2)-(1)(0)]-3[(1)(-2)-(1)(1)]+1[(1)(0)-(1)(1)]

=1(-2-0)-3(-2-1)1(0-1)

=1(-2)-3(-3)1(-1)

= -2+ 9- 1

Dy=6

______________________________________

Step4: we have to calculate Dz!!

$\: \: \: \: \: \: \: \: \: \: \: |1 \: \: \: \: 1 \: \: \: \: \: \: \: 3| \\ d \: = |1 \: - 1 \: \: \: 1| \\ \: \: \: \: \: \: \: \: \: \: \: | \: \: 1 \ \: \: 1 \: \: \: \: \: \: 0|$

=1[(-1)(0)-(1)(1)] -1[(1)(0)-(1)(1)] +3[(1)(1)-(-1)(1)]

=1(0-1) -1(0-1) +3(1+1)

=1(-1)-1(-1)3(2)

=-1+1+6

Dz=6

______________________________________

Now, By Cramer's Rule~~

$x = \frac{dx}{x} = \frac{6}{6} = 1$

$y = \frac{dy}{y} = \frac{6}{6} = 1$

$z = \frac{dz}{z} = \frac{6}{6} = 1$

x=1 y=1 z=1!!

Hope it helps!!