Problem to solve
Solve \[x-y+3z=2\], \[2x-3y=-4\], \[x+y+z=-2\]
Solution
STEP 1
Solve the following system:
{x - y + 3 z = 2
-3 y + 2 x = -4
x + y + z = -2
STEP 2
Express the system in matrix form:
(1 | -1 | 3
2 | -3 | 0
1 | 1 | 1)(x
y
z) = (2
-4
-2)
STEP 3
Solve the system with Cramer's rule:
x = 2 | -1 | 3
-4 | -3 | 0
-2 | 1 | 1/1 | -1 | 3
2 | -3 | 0
1 | 1 | 1 and y = 1 | 2 | 3
2 | -4 | 0
1 | -2 | 1/1 | -1 | 3
2 | -3 | 0
1 | 1 | 1 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/1 | -1 | 3
2 | -3 | 0
1 | 1 | 1
STEP 4
Evaluate the determinant 1 | -1 | 3
2 | -3 | 0
1 | 1 | 1 = 14:
x = 2 | -1 | 3
-4 | -3 | 0
-2 | 1 | 1/14 and y = 1 | 2 | 3
2 | -4 | 0
1 | -2 | 1/14 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/14
STEP 5
Evaluate the determinant 2 | -1 | 3
-4 | -3 | 0
-2 | 1 | 1 = -40:
x = (-40)/14 and y = 1 | 2 | 3
2 | -4 | 0
1 | -2 | 1/14 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/14
STEP 6
The gcd of -40 and 14 is 2, so (-40)/14 = (-20×2)/(7×2) = -20/7:
x = -20/7 and y = 1 | 2 | 3
2 | -4 | 0
1 | -2 | 1/14 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/14
STEP 7
Evaluate the determinant 1 | 2 | 3
2 | -4 | 0
1 | -2 | 1 = -8:
x = (-20)/7 and y = (-8)/14 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/14
STEP 8
The gcd of -8 and 14 is 2, so (-8)/14 = (-4×2)/(7×2) = -4/7:
x = (-20)/7 and y = -4/7 and z = 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2/14
STEP 9
Evaluate the determinant 1 | -1 | 2
2 | -3 | -4
1 | 1 | -2 = 20:
x = (-20)/7 and y = (-4)/7 and z = 20/14
STEP 10
The gcd of 20 and 14 is 2, so 20/14 = (10×2)/(7×2) = 10/7:
Answer: | 
 | x = (-20)/7 and y = (-4)/7 and z = 10/7