Solving Linear Systems with Cramer's Rule

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lina29
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Homework Statement


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


Homework Equations



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

y1 3 1
y2 a 0
y3 a 1



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[itex]^{}2[/itex]-5a
and x1=(ay1+(-3+a)y2-ay3)/a[itex]^{}2[/itex]-5a

But that was counted wrong. I don't understand where I went wrong since I have doublechecked my calculations
 
on Phys.org
Hi lina29! :smile:

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
 
I'm sorry I typed the equation wrong it was
-5x1+ ax2+x3. Even using -a2+5a as the denominator I got it wrong
 
and when doing x2=(a-5)y2/(-a2+5a) I also got it counted wrong. Is there something I'm missing out?
 
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
 
Could you explain to me where I'm going wrong?
 
lina29 said:
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

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

Here's for instance the formula from wikipedia:
b9845d679f1ad3b093e3572a8120f8d6.png
 
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
 
lina29 said:
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

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