Linear Algebra: Inverse of Elementary Matrix

ephemeral1
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Homework Statement



What is the inverse of this elementary matrix? 2x2 matrix
1 0
3 -1


E=

Homework Equations



None

The Attempt at a Solution


This is what I think:
1 0
-3 -1

Is that right? Or do you leave the 3 positive? If so, please explain. Thank you.
 
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ephemeral1 said:

Homework Statement



What is the inverse of this elementary matrix? 2x2 matrix
1 0
3 -1


E=

Homework Equations



None

The Attempt at a Solution


This is what I think:
1 0
-3 -1

Is that right? Or do you leave the 3 positive? If so, please explain. Thank you.

Do you get the identity matrix when you multiply your inverse by the original?
 
No, I did not. But when I multiply
1 0
3 -1 by the original, I get the identity matrix. I don't understand why the 3 is positive. Don't we have to change the sign on additional constant? I thought the 3 is the additional constant? Or is it because the 1 on the second row is negative, we don't have to change the sign on the -3 when doing the inverse? Please explain. Thank you.
 
What method did you use to calculate the inverse? That should answer your question.
 
you are correct you must change the sign of 3, however the determinant is equal to -1. remember you have to divide by the determinant when using cramer's rule.
 
another way to calculate the inverse is to write the matrix equation AX=I and solve for the unknowns in X.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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