Finding k to Make Matrix Singular: Solving the Mystery

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Homework Help Overview

The discussion revolves around finding a value of k that will make a given 3x3 matrix singular. The original poster attempts to calculate the determinant of a minor instead of the full matrix to determine the conditions for singularity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants question the validity of using a minor's determinant to conclude about the singularity of the entire matrix. There is a suggestion to calculate the determinant of the full matrix instead.

Discussion Status

Participants have provided guidance on the necessity of calculating the determinant of the entire matrix rather than just a minor. The original poster acknowledges a misunderstanding and expresses gratitude for the clarification.

Contextual Notes

The original poster's approach was based on a misinterpretation of how determinants relate to matrix singularity, leading to an incorrect submission.

mr_coffee
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Hello everyone I'm suppose to find a k, that will make this matrix singular, well i looked up what that ment, and it means i have to make the determinant 0. So i the matrix is:
-3 2 1
-9 8 3
(9+k) 2 0

so i found a random minor, say M13..
so i would have
-9 8
(9+k) 2

det = (-9)(2) - [8(9+k)] = -18 -72 -8k = -90 -8k;
so i want k to be 0, so i solve for k
-90 -8k = 0;
k = -90/8, i submitted it, and it was incorrect, any ideas why?
 
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Why would making one of the minors 0 make the whole determinant 0? Did you even calculuate the determinant of your matrix? Basically, what you've done is taken:

a*det(A) - b*det(B) + c*det(C)

and said that if as long as one of det(A), det(B), or det(C) is zero, the whole thing is somehow zero. Clearly that's wrong. If the determinants of all the minors were zero, then you'd be fine, but you can't just show one of them to be zero and think that it does anything.

A matrix is singular when its rows and/or columns are linearly dependent. The easiest way to do this is to make two rows or two columns scalar multiples of each other.
 
You need to work out the determinant of the entire matrix, not just one of its minors.
 
The determinant to be zero is the whole determinant. The det of A.

Work out the det of A with the elements given and equate that to zero.
 
ohh my bad, i don't know why I thought that, too much coffee. Thanks it worked out great!
 

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