Question

An nn matrix A is invertible if and only if (choose ALL correct answers)
A. A is a product of elementary matrices
B. Rank A=n
C. AX=0 has a unique solution where X is an n1 matrix
D. A can be carried to the identity matrix I by a sequence of row operations
E. XA=0 has a unique solution where X is an 1n matrix

Answer 1

is there any one here??

Answer 2

I know it's true for A, B, and D. I'm not sure about C or D.

Answer 3

well u are not sure about C and E, me too

Answer 4

I tend to think C is true, and if that's true, it only makes sense that E is also true.

Answer 5

I revise my statement. C is definitely true.

Answer 6

And it would then make absolutely no sense for E to be false then.

So they're all correct.

Answer 7

yup they are all true
thanks :)

Answer 8

you're welcome.

Answer 9

A-D are true, unsure about E.

Answer 10

If A is invertible then E is true.
If E is true, one has to show that A is invertible.
If A is not invertible there is a row of it that is a linear combinations of all the others.
To fix the idea, suppose that that row is the first one. Hence
\[
r_1= a_2 r_2 + a_3 r_3+ \cdots a_n r_n
\]
where all the a_i are not all zeros. Now take
\[
X=(-1, a_2, a_3, \cdots , a_n)\ne {\bf 0} \text { but}\\
X A = {\bf 0}
\]
So E is violated.