Question

Which of the following statements are true?

Answer 1

I know A, E and F are correct but what about the others. Are any of them correct as well?

Answer 2

Linear ?

Answer 3

yes, A, E, F are true. That's correct.

Answer 4

before I type, I want to confirm you are there. Are you?

Answer 5

yep

Answer 6

Let's look at the following: (Columns > Rows)
\(\color{#000000}{ \displaystyle A=\left[\begin{matrix}a_{(1),1} & a_{(2),1}& a_{(3),1} \\ a_{(1),2} & a_{(2),2}& a_{(3),2}\end{matrix}\right] }\)

Then, you are going to have a free variable, and your
equation should be not only consistent, but in fact it
would have infinitely many solutions. \(\bf \color{red}{Unless}\) you end
up with something like \(0=6\), e.g.
\(\color{#000000}{ \displaystyle \left[\begin{matrix}1 & 0& 2& 4 \\ 0 & 0& 0& 6\end{matrix}\right] }\) where \(\color{#000000}{ \displaystyle \left[\begin{matrix} 4 \\ 6\end{matrix}\right]={\bf b} }\)

and such case when you get 0=non-zero, is the only case.
Notice that you don't have a pivot in every row of A. (and can't when columns>rows, and Ax=b is inconsistent).

Now, what if you had: (Columns < Rows)
\(\color{#000000}{ \displaystyle A=\left[\begin{matrix}a_{(1),1} & a_{(2),1} \\ a_{(1),2} & a_{(2),2} \\ a_{(1),3} & a_{(2),3} \end{matrix}\right] }\)

Then, even if you can reduce it to something like:
\(\color{#000000}{ \displaystyle A=\left[\begin{matrix}1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix}\right] }\)

and can the pivot be right above a pivot?

Answer 7

can hardly connect, sorry.

Answer 8

that was for B.

Answer 9

I will stop using transcripts, they are very complicated and not well readable.... anyway, #3 ...

Answer 10

Statement C is really a definition:

Consider something like: \(\color{#000000}{ A{\bf x}={\bf b} }\)
Then, (let's look in \(\mathbb{R}^3\) case).

Where \(\color{#000000}{ \displaystyle A=\left[\begin{matrix}a & d& g\\ b & e&h\\ c & f&i\end{matrix}\right] }\), \(\color{#000000}{ \displaystyle {\bf b}=\left[\begin{matrix}a \\ d \\ g \end{matrix}\right] }\) and \(\color{#000000}{ \displaystyle {\bf x}=\left[\begin{matrix}x_1 \\ x_2 \\ x_3 \end{matrix}\right] }\)

then if you were to un-setup the augmented matrix A, you get:
\(\color{#000000}{ \displaystyle \left[\begin{matrix}a \\ b \\ c \end{matrix}\right] x_1 +\left[\begin{matrix}d \\ e \\ f \end{matrix}\right] x_2 +\left[\begin{matrix}g \\ h \\ i \end{matrix}\right]x_3 =\left[\begin{matrix}k \\l \\ m \end{matrix}\right] }\)

and after re-writing it like this, it should be clear that a solution to the system is by definition a linear combinations of columns of A, multiplied by the solutions for \(x_i\).

Answer 11

You can take a homogeneous equation, that is consistent (at least for trivial solution), regardless!

Let, \(\color{#000000}{ \displaystyle \left[A~{\bf b}\right]=\left[\begin{matrix}1 & 0&0& 0 \\ 0 & 1& 0& 0\\ 0 & 0& 1& 0\end{matrix}\right] }\)

Does every row contain a pivot?
Is this equation consistent?


(Also, you can try any b in fact ... )

Answer 12

and the reason that you can try any b, is because the columns of A (by definition) span \(\mathbb{R}^3\) (or any vector \(b\) that can possibly be in \(\mathbb{R}^3\)), since \(A=I_3\).

Answer 13

so, other than A, E and F, C and D is also correct, right?
I asked my instructor about the question and he told me that A is wrong and that C is correct along with E and F. I submitted what he said but that was wrong too.