Solving Linear Systems with Cramer's Rule

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Homework Help Overview

The discussion revolves around solving a system of linear equations using Cramer's Rule, specifically focusing on the conditions under which the determinant of the coefficient matrix is non-zero. Participants are analyzing the calculations of determinants and their implications for solving the system.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate the determinants of matrices A and A1, with varying results leading to confusion about the correctness of their calculations. Questions about sign errors and the correct application of determinant formulas are raised.

Discussion Status

There is an ongoing examination of the determinant calculations, with some participants suggesting corrections to others' work. Multiple interpretations of the determinant formulas are being explored, and while some guidance has been offered, there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There are indications of miscommunication regarding the setup of the matrices and the calculation methods for determinants.

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^{}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
 
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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
 
Ah, but with your correction your det(A) is still wrong.
Can you recalculate it?
 
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
 
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.
 
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
 
  • #10
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
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
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).
 
  • #13
Got it! (a+5)y2
Thank you so much!
 
  • #14
You're welcome! :smile:
 

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