Question

suppose that b is row equivalent to the nxn invertible matrix a. show that b is invertible.

Answer 1

Allowed to use determinants?

Answer 2

Well we haven't covered determinants yet so no.

Answer 3

An invertible matrix is row equivalent to the identity matrix with n-rows, right?

Answer 4

yes

Answer 5

Well then, if b is row equivalent to a, then there exists a sequence of elementary row operations to turn b into a, right?

Answer 6

yes, unfortunately my brain can't see it yet.

Answer 7

Well then, there exists a sequence of elementary row operations to turn b into a, and in turn, there exists a sequence of elementary row operations to turn a into the identity matrix. Therefore, there exists a sequence of elementary row operations to turn b into the identity matrix, thus, b must be invertible.

Answer 8

Okay I see what you are saying, I'm just not sure on how to show that.

Answer 9

Okay..., let me put it in another way...
A matrix is invertible if and only if it is row equivalent to the identity matrix I.
a is invertible, therefore, there exist some elementary matrices
E[1] E[2] ... E[n] such that
E[1] E[2] ... E[n] a = I.

Now, b is row equivalent to a. Therefore, there exist elementary matrices
F[1] F[2] ... F[m] such that
F[1] F[2] ... F[m] b = a
means we can multiply E[1] E[2] ... E[n] to both sides of the equation, we get
E[1] E[2]...E[n] F[1] F[2]...F[m] b = E[1] E[2] ... E[n] a = I
Thus, b is row equivalent to the identity matrix, meaning it's invertible.

Answer 10

By the way, write
\[\large E[n] = E_n\]
my computer is just lagging right now...