Why Are These Matrices Not Invertible?

[SiddharthThakur]

Question
a) For each of the following matrices explain why the matrix is not invertible.
i)[2,0,0_0,0,0_9,3,0]

ii)[3,4,-6_7,2,1_-6,-812]

iii)[5,-2,15_1,-4,3_2,1,6]
b) Suppose A is an n×n matrix with the property that the equation Ax = 0 has only the trivial
solution. Without using the Invertible Matrix Theorem, explain directly why the equation
Ax = b must have a solution for each b in Rⁿ

 

[Jameson]

Hi SiddharthThakur,

A couple of things need to be said.

1) We don't do homework for people. You can't post a bunch of problems without any effort and expect us to solve them. We want to help you know how to them though and are all very happy to do them with you but not for you.

2) It appears that you have an assignment that is for a grade and since we don't know if you are allowed to receive help on it, I can't allow you to get more help on it until you address this issue. I will let you respond to this in one of your threads but if you can't show that you are allowed to get help on this assignment then I will have to close the threads. It's our policy to help prevent cheating.

Jameson

 

[Jameson]

Ok I reread the PDF and it seems you can receive help in a way, but you must cite the sources. So as long as no one here gives you any answers and you agree to cite MHB as a source that helped you.

 

[SiddharthThakur]

Ok..

 

[Jameson]

For i, ii and ii in part a what have you tried? Here is the invertible matrix theorem for your reference. Looking at the pivot positions and the determinant is a good place to start.

For b, use the rank nullity theorem. \(\displaystyle \text{dim Col A}+ \text{ dim Null}=n\). Which part of that are you given already?

 

[Jameson]

Since the deadline for the assignment in question has passed, I will make some more comments on solving the problem so others can use the information in the future.

i) $A= \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right)$. We can quickly see that $\text{det }A=0$, thus by the Invertible Matrix Theorem $A$ is not invertible.

ii) $B= \left( \begin{array}{ccc} 3 & 4 & -6 \\ 7 & 2 & 1 \\ -6 & -8 & 12 \end{array}\right)$.

If we notice that $R_3=2R_1$ then $R_3$ is row equivalent to $[0,0,0]$ and has less than 3 pivot positions, thus is not invertible.

iii) $C= \left( \begin{array}{ccc} 5 & -2 & 15 \\ 1 & -4 & 3 \\ 2 & 1 & 6 \end{array}\right)$

This takes a bit more calculation but if we row reduce the matrix to echelon form then we see that again there isn't a pivot position in every row, thus the matrix is not invertible.

(b) The rank-nullity theorem says that $\displaystyle \text{dim Col A}+ \text{ dim Null}=n$. Since there is only the trivial solution to $A \vec{x}=\vec{0}$, $\text{ dim Null}=0$ which implies that $\text{dim Col A}=n$. Since $A$ is an $n$ by $n$ matrix with $n$ linearly independent column vectors, it spans $\mathbb{R}^n$.

 

[The Chaz]

Just a note on (iii):

$R_1 = -2(R_3) - (R_2)$
So the rows are linearly dependent.

 

[Jameson]

Just a note on (iii):

$R_1 = -2(R_3) - (R_2)$
So the rows are linearly dependent.

I just looked for one row being a scalar multiple of another row. Nice catch! :) I don't have a great eye for that sort of thing all the time.