Linear Algebra and rank problem.

georgeh
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I have the following problem which I can't figure out.

Let A = [a_11,a_12;a_13; a_21; a_22; a_23;]
Show that A has rank 2 if and only if one or more of the determinants

| a_11,_a_12; a_21,a_22| , |a_11,a_13;a_21,a_23|,|a_12,a_13;a_22,a_23|
I know its a 2x3 matrix..which the det. wouldn't apply since it is not square. Not sure how to proceede
 
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You might start by stating the problem correctly

"Show that A has rank 2 if and only if one or more of the determinants
| a_11,_a_12; a_21,a_22| , |a_11,a_13;a_21,a_23|,|a_12,a_13;a_22,a_23|" is what?? What is supposed to be true about them?

"I know its a 2x3 matrix..which the det. wouldn't apply since it is not square."
That's irrelevant- the problem doesn't say anything about the determinant of A (which doesn't exist) only the determinants of those 2 by 2 subsets.
 
HallsofIvy said:
You might start by stating the problem correctly

"Show that A has rank 2 if and only if one or more of the determinants
| a_11,_a_12; a_21,a_22| , |a_11,a_13;a_21,a_23|,|a_12,a_13;a_22,a_23|" is what?? What is supposed to be true about them?
the det is not equal to zero..
"I know its a 2x3 matrix..which the det. wouldn't apply since it is not square."
That's irrelevant- the problem doesn't say anything about the determinant of A (which doesn't exist) only the determinants of those 2 by 2 subsets.
Yeah i states how they had it though. sorry.
they said.. the det != 0
 
Okay, now that we have that straightened out, exactly what is your definition of "rank"? What do you get if you "row reduce" A?
 
the rank is the dimensions of the row space and column space of a matrix.
So when we do r-r-e, wherever we get leading ones, that is the rank #.
 
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