Matrix Inversion Problem

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  • #1

Homework Statement


Let A be a 2x3 matrix with real coefficients, and suppose that neither of the two rows of A is a scalar multiple of the other. By quoting an appropriate theorem from linear algebra, show that for some j [tex]\in[/tex] {1,2,3}, the 2x2 matrix obtained by omitting the jth column of A is invertible


Homework Equations


Not really any necessary equations that I can think of. Only one I suppose is:

AA[tex]^{-1}[/tex] = I
So A has to be a square matrix in order for it to be invertible.

The Attempt at a Solution


I know in order for a matrix to be invertible, it has to be an nxn matrix. Also, the determinant of A cannot be equal to 0. So, if the rows are not multiples of each other, does that automatically mean that any 2x2 submatrix of A have a determinant not equal to 0?

Any help would be appreciated!
 

Answers and Replies

  • #2
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I know in order for a matrix to be invertible, it has to be an nxn matrix. Also, the determinant of A cannot be equal to 0. So, if the rows are not multiples of each other, does that automatically mean that any 2x2 submatrix of A have a determinant not equal to 0?

Write it out and solve the equations: if [tex]\det \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \neq 0[/tex], what does that tell you about the relationship between [tex](a \,\, b)[/tex] and [tex](c \,\, d)[/tex]?
 
  • #3
I know then that ad-bc[tex]\neq[/tex]0. But I don't know where to go after that. :( I think I have to use the Gram-Schmidt process, but I'm not sure. Help please?
 
  • #4
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You can deduce from whether [tex]ad - bc = 0[/tex] whether there exists any [tex]\beta[/tex] such that [tex](a\,\,b) = \beta(c\,\,d)[/tex] or [tex]\beta(a\,\,b) = (c\,\,d)[/tex]. Think about the ratios [tex]c/a[/tex] and [tex]d/b[/tex], or their inverses if one of [tex]a[/tex] and [tex]b[/tex] is zero.
 
  • #5
The ratios should be equal to the constant [tex]\beta[/tex], but that's assuming that ad-bc=0. If that was the case, then that matrix wouldn't be invertible. Am I misunderstanding something?
 
  • #6
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My point is that you can prove (and you should) that this implication goes in both directions: one row is a constant multiple of the other if and only if [tex]ad - bc = 0[/tex].
 
  • #7
Wow, that's actually pretty obvious. If the rows aren't scalar multiples of each other, than the determinant can't possible equal to 0. And if the determinant isn't equal to 0, then the matrix is invertible!

One last problem, do we have to prove anything about omitting a column? I thought that was the brunt of the problem, and that was what was throwing me off.

Thanks for all the help so far!
 
  • #8
352
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Well, you need to take the fact that neither row of the original matrix [tex]A[/tex] is a scalar multiple of the other, and conclude from that that one of the three possibilities after a column is omitted still has this property. There is a small argument you need to make there.
 
  • #9
I can't think of how to prove that any 2x2 matrix left over by omitting one of the columns will be guaranteed to be non-scalar multiples. The matrix I have in mind is:

[tex]
\left( \begin{array}{cc} 3 4 7 \\ 6 8 8 \\ \end{array} \right) \
[/tex]

If I omit column j=3, I'll have a 2x2 matrix with rows that are scalar multiples. Since they are scalar multiples, they're determinant is equal to 0, and thus not invertible. If they're not invertible, then it doesn't work by omitting ANY column, but only a column that would leave a matrix that have no scalar rows.
 

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