Step by step
The question
Question: solve x + 2z = 2, x + 2y + z = −2, x − y − z = −3 by Cramer's rule.
Three equations, three letters: x, y and z.
Cramer's rule, like with two equations: x is D x over D, y is D y over D, z is D z over D.
Four determinants, each by the diagonals.
1 Find D
D: the numbers in front of x, y and z, in their places.
We write the first two columns again, on the right.
Then every diagonal has three numbers.
Down: 1 × 2 × (−1), 0 × 1 × 1, 2 × 1 × (−1).
−2, 0, −2.
Together: −4.
Up: 1 × 2 × 2, −1 × 1 × 1, −1 × 1 × 0.
4, −1, 0.
Together: 3.
Down minus up: −4 minus 3 is −7.
So D = −7.
2 Find Dx
D x: the x column is replaced by the right sides, 2, −2, −3.
Down: 2 × 2 × (−1), 0 × 1 × (−3), 2 × (−2) × (−1).
−4, 0, 4.
Together: 0.
Up: −3 × 2 × 2, −1 × 1 × 2, −1 × (−2) × 0.
−12, −2, 0.
Together: −14.
Down minus up: 0 minus (−14) is 14.
So D x = 14.
3 Find Dy
D y: the y column is replaced by the right sides, 2, −2, −3.
Down: 1 × (−2) × (−1), 2 × 1 × 1, 2 × 1 × (−3).
2, 2, −6.
Together: −2.
Up: 1 × (−2) × 2, −3 × 1 × 1, −1 × 1 × 2.
−4, −3, −2.
Together: −9.
Down minus up: −2 minus (−9) is 7.
So D y = 7.
4 Find Dz
D z: the z column is replaced by the right sides, 2, −2, −3.
Careful!
For D z we replace the z column, the third one.
The other two stay.
Down: 1 × 2 × (−3), 0 × (−2) × 1, 2 × 1 × (−1).
−6, 0, −2.
Together: −8.
Up: 1 × 2 × 2, −1 × (−2) × 1, −3 × 1 × 0.
4, 2, 0.
Together: 6.
Down minus up: −8 minus 6 is −14.
So D z = −14.
5 Divide
x is D x over D: 14 over (−7) is −2.
y is D y over D: 7 over (−7) is −1.
z is D z over D: −14 over (−7) is 2.
So the answer is (−2, −1, 2).
6 Check
Let's check in every equation.
We put x = −2, y = −1, z = 2 into ①.
We get 2, the same as the right side.
Now ②.
We get −2, the same as the right side.
Now ③.
We get −3, the same as the right side.
All of them work.
So the answer is right!
Let's look back.
First: find D.
Then: find Dx.
Then: find Dy.
Then: find Dz.
Then: divide.
Last: check.