Linear Algebra Matrix Basic Questions

[JJBladester]

Homework Statement

True or False (correct any false statement)

D) If $$A_{nxn}$$ is row equivalent to $$I_{n}$$, A has rank n.

I) If A =

1 -1 3
0 -1 4
0 0 -1

AX=0 has trivial solution only.

J) For matrix A above (I), AX=X has infinitely many solutions.

N) If AX=B has exactly one solution, A is invertible.

Homework Equations

N/A

The Attempt at a Solution

D) I understand rank to mean the number of non-zero rows in a matrix, thus if there are no non-zero rows, then rank = n for a nxn matrix.

Saying that $$A_{nxn}$$ is invertible means that there exists an nxn matrix B such that AB=BA=$$I_{n}$$. So, invertible means that you were able to perform enough row operations on A to get it to become the inverse, so I think that the answer is TRUE.

I) Not sure where to start on this one.

J) Not sure where to start on this one.

N) True... because the inverse of a matrix is unique, there can only be one solution to AX=B.

 

[Office_Shredder]

Not sure what row equivalent means, but for D, the rank is the dimension of the span of the rows of the matrix (one equivalent definition at least, which seems to be the one you're using)... so for example the matrix

[1 1 1]
[1 1 1]
[1 1 1]

has rank 1, even though all its rows are non-zero

For (I), is A invertible? Invert it! Actually, you don't need to actually calculate the inverse to solve A^-10, nor do you need to calculate the inverse to prove A is invertible

For (J), you can calculate whether 1 is an eigenvalue (if you know what that means). Easier and more intuitively, let X=(x,y,z). then AX = X gives you three equations in three unknowns (x,y,z). See what you can find out about it

You did (N) backwards: you showed if A is invertible, then AX=B has a unique solution. Fair enough, but that's not what the question asked