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August 29th, 2011, 08:23 PM   #1
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Invertible matrix

Let A be an nxn matrix. Prove that if A is not invertible, then there exists an nxn matrix B such that but
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August 29th, 2011, 09:05 PM   #2
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Re: Invertible matrix

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Originally Posted by problem
Let A be an nxn matrix. Prove that if A is not invertible, then there exists an nxn matrix B such that but
A not invertible means (among other things) that the determinant of A is zero.

If A is the zero matrix, then we are done.

Otherwise, suppose that A is not the zero matrix, yet det(A) = 0.

Then A has at least one non-zero entry, say A_ij = c, for some value of c, and some 0 <= i,j <= n

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August 29th, 2011, 09:24 PM   #3
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Re: Invertible matrix

The determinant of a matrix will be introduced after the section where this question is from. Aside from determinant, can we use any property of invertible matrix? Thanks for the help.
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August 31st, 2011, 06:30 AM   #4
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Re: Invertible matrix

If an n by n maA is not "invertible", then it is not "one-to-one". That is, there exist two distinct vectors, X and Y, such that AX= AY. Of course, AX- AY= A(X- Y)= 0.
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