Problem to solve
Solve \[-\frac{3x}{2}+3y+z=2\], \[3x-y+z=5\], \[\frac{x}{3}-y+z=2\]
Solution
STEP 1
Solve the following system:
{-(3 x)/2 + 3 y + z = 2
3 x - y + z = 5
x/3 - y + z = 2
STEP 2
Express the system in matrix form:
(-3/2 | 3 | 1
3 | -1 | 1
1/3 | -1 | 1)(x
y
z) = (2
5
2)
STEP 3
Solve the system with Cramer's rule:
x = 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1/-3/2 | 3 | 1
3 | -1 | 1
1/3 | -1 | 1 and y = (-3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/(-3/2 | 3 | 1
3 | -1 | 1
1/3 | -1 | 1) and z = (-3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/(-3/2 | 3 | 1
3 | -1 | 1
1/3 | -1 | 1)
STEP 4
Evaluate the determinant -3/2 | 3 | 1
3 | -1 | 1
1/3 | -1 | 1 = (-32)/3:
x = 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1/(-32/3) and y = (-3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/(-32/3) and z = (-3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/(-32/3)
STEP 5
The gcd of 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1 and -32/3 is 1/3, so 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1/((-32)/3) = (3 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1)/(3 ((-32)/3)) = -3/32 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1:
x = -3/32 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1 and y = (-3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/((-32)/3) and z = (-3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/((-32)/3)
STEP 6
The gcd of -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1 and -32/3 is 1/3, so (-3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/((-32)/3) = (3 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/(3 ((-32)/3)) = -3/32 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1:
x = (-3 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1)/32 and y = -3/32 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1 and z = (-3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/((-32)/3)
STEP 7
The gcd of -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2 and -32/3 is 1/3, so (-3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/((-32)/3) = (3 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/(3 ((-32)/3)) = -3/32 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2:
x = (-3 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1)/32 and y = (-3 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/32 and z = -3/32 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2
STEP 8
Evaluate the determinant 2 | 3 | 1
5 | -1 | 1
2 | -1 | 1 = -12:
x = (-3)/32×-12 and y = (-3 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/32 and z = (-3 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/32
STEP 9
(-3)/32 (-12) = 9/8:
x = 9/8 and y = (-3 -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1)/32 and z = (-3 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/32
STEP 10
Evaluate the determinant -3/2 | 2 | 1
3 | 5 | 1
1/3 | 2 | 1 = (-11)/2:
x = 9/8 and y = (-3)/32×-11/2 and z = (-3 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/32
STEP 11
(-3)/32×(-11)/2 = 33/64:
x = 9/8 and y = 33/64 and z = (-3 -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2)/32
STEP 12
Evaluate the determinant -3/2 | 3 | 2
3 | -1 | 5
1/3 | -1 | 2 = (-137)/6:
x = 9/8 and y = 33/64 and z = (-3)/32×-137/6
STEP 13
(-3)/32×(-137)/6 = 137/64:
Answer: | 
 | x = 9/8 and y = 33/64 and z = 137/64