Prove: A 2x2 matrix is nonsingular if and only if the determinant != 0

  • Thread starter superdave
  • Start date
  • #1
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Homework Statement




Prove: A 2x2 matrix is nonsingular if and only if the determinant != 0

The Attempt at a Solution



I need to prove this, using logic and maybe the theorem that a n x n matrix is nonsingular if and only if it is row equivalent to I_n.

I could use a push in the right direction. I need to show a proof for both ways, as it is an if and only if statement.
 

Answers and Replies

  • #2
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I'm no linear algebra expert, but I would start by writing that if the matrix is written:
a b
c d

Then it's singular if a = x*b and c=x*d (or a=x*c and b=x*d), and work my way to the formula for the determinant.
 
  • #3
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Then it's singular if a = x*b and c=x*d (or a=x*c and b=x*d)

Ok, but where does this come from?
 
  • #4
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It would come from the fact that if the matrix is singular, then column one a multiple of the other (or one row is a multiple of the other). Maybe this isn't how it's supposed to be done though, so you might be better off waiting for somebody else's ideas than going off mine :/
 
  • #5
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For one direction, assume matrix A is invertible. Then there is matrix B such that AB=1, where 1 is identity matrix. Take the determinant of both sides of AB=1.

For the other direction, assume det A is invertible. Using the determinant and cofactor expansions, you can find a nice form for the inverse of A
 
  • #6
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I probably should have mentioned. I don't have the definition of the determinant to work with.

the problem actually says Show that the 2 x 2 matrix A is nonsingular if and only if ad-bc != 0.

I've figured out the If matrix A is nonsingular, then ab-bc != 0 side. I just need the if ad - bc != 0, Matrix A is nonsingular side.
 
Last edited:
  • #7
614
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Given matrices of the form
a b = A
c d

and

d -b = B
-c a

How can you ensure AB=1?
 

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