Cramer's Rule Example 3x3 Matrix

This worksheet help you to understand how to find the unknown variables in linear equation. In this example We are going to find three unknown variables from three linear equations.

Solve the following equation and find the value of x, y, z.
3x + y + z = 3
2x + 2y + 5z = -1
x - 3y - 4z = 2

Δ =
3 1 1
 2 2 5
 1-3-4

= 3(-8 + 15) - 1(-8 - 5) + 1(-6 - 2)
= 21 + 13 - 8
= 26

Here Determinant is not equal to zero, Hence Cramer's rule can be applicable.
X = ∆x / ∆ , y = ∆y / ∆ , z = ∆z / ∆
Δx =
 3 1 1
-1 2 5
 2-3-4

=3( -8 + 15) - 1(4 - 10) + 1(3 - 4)
= 21 + 6 - 1
x= 26


Δy =
 3 3 1
 2-1 5
 1 2-4

= 3(4 - 10) - 3(-8 - 5) + 1(4 + 1)
=-18 + 39 + 5
y= 26

Δz =
 3 1 3
 2 2-1
 1-3 2

= 3(4 - 3) - 1(4 + 1) + 3(-6 -2)
= 3 - 5 - 24
z= -26

By using Cramer's rule:
x = ∆1 / ∆
x = 26 / 26
x = 1
y = ∆2 / ∆
y = 26 / 26
y = 1
z = ∆3 / ∆
z = 26 / -26
z = -1
Therefore the required x = 1, y = 1 & z = -1

When you try this calculation on your own, use this Cramer's Rule Calculator to verify your answers.

Cramer's Rule Practice Problems:
Solve the following equations using Cramer's Rule
  1. 4x+3y+1=0 and -y+2x=3
  2. 1/y+2/x=5 and -1/x+3/y=8
  3. x + 2y - z = 5
    2x - y + z = 1
    3x + 3y - 2z = 8
  4. x + z = 1
    y + z = 1
    x + y = 4
  5. 4x - y + 3z = 2
    x + 5y - 2z = 3
    3x +2y + 4z = 6