# Linear Algebra Cramer's Rule

1. Sep 26, 2011

### lina29

1. The problem statement, all variables and given/known data
Solve the system:
ax1+3x2+x3= y1
ax2= y2
-5x1+ax2-x3=y3

using cramer's rule for those a such that the det of A is not 0

2. Relevant equations

x1= det(A1)/det(A)
det(A1)=a11+c11+a21c21+a31c31
and A1=

y1 3 1
y2 a 0
y3 a 1

3. The attempt at a solution
By using those equations I got the det(A1) to go ay1+(-3+a)y2-ay3
and the det(A) was a$^{}2$-5a
and x1=(ay1+(-3+a)y2-ay3)/a$^{}2$-5a

But that was counted wrong. I don't understand where I went wrong since I have doublechecked my calculations

2. Sep 26, 2011

### I like Serena

Hi lina29!

You appear to have made a few mistakes with signs.
For starters, det(A) is not a2-5a, but -a2+5a.
How did you calculate it?

And your A1 should be (note the -1 in the lower right corner):
y1 3 1
y2 a 0
y3 a -1

3. Sep 26, 2011

### lina29

I'm sorry I typed the equation wrong it was
-5x1+ ax2+x3. Even using -a2+5a as the denominator I got it wrong

4. Sep 26, 2011

### I like Serena

Can you recalculate it?

5. Sep 26, 2011

### lina29

and when doing x2=(a-5)y2/(-a2+5a) I also got it counted wrong. Is there something I'm missing out?

6. Sep 26, 2011

### lina29

I'm still getting a2-5a. Using the equation a11c11+a12c12+a13c13.
a11=a
a12=3
a13=1
c11=a
c12=0
c13=-5a

and combining them together I get a2-5a

7. Sep 26, 2011

### I like Serena

Yes, your det(A) is wrong, so x2 is wrong too.
You appear to have made the same mistake calculating det(A2).
Strangely you did calculate det(A1) properly.

8. Sep 26, 2011

### lina29

Could you explain to me where I'm going wrong?

9. Sep 26, 2011

### I like Serena

I don't know this equation for a determinant, but it is not right.

Here's for instance the formula from wikipedia:

10. Sep 26, 2011

### I like Serena

Oh, I think your c11,c12,c13 are the determinants of the sub matrices in the lower 2 rows.
In that case you miscalculated c13.

11. Sep 26, 2011

### lina29

c11, c12, and c13 are determinants of the submatrices. In class we were taught to calculate the determinant using cofactors. Using your method though I had det(A2)= (a-5)y2 -ay3 and det(A)=a2+5a

12. Sep 26, 2011

### I like Serena

Yep, I got that at second sight.
See my previous post.

So here you have the right det(A), but you still miscalculated det(A2).

13. Sep 26, 2011

### lina29

Got it! (a+5)y2
Thank you so much!

14. Sep 26, 2011

### I like Serena

You're welcome!