Question

How do I write the matrix
[1 0 0 | 0]
[0 1 0 | 0]
[0 0 0 | 0]

as a basic solution? or parametric form? (This is a 3x3 matrix btw)

Answer 1

Do you know what basic solution or parametric form would look like? Not for this particular matrix, but in general.

Answer 2

yes :D, I'm just confused but the column of 0's

Answer 3

Can you give me the general forms, because I don't completely remember them.

Answer 4

I mean, I know how to simplify it, just not the names for it.

Answer 5

\[
x=0\\
y=0\\
z=t
\]Basically, \(x\) and \(y\) must be \(0\), while \(z\) can be whatever you want it to be.

Answer 6

Another way to write this is: \[
\begin{bmatrix}
0\\0\\1
\end{bmatrix}t
\]

Answer 7

parametric is like: (An example)
x1 = (9/5)s - (14/5)t
x2 = (-4/5)s - (1/5)t
x3 = s
x4 = t

basic solution is like:
x1[s] = [9/5]
[-4/5]
[1]
[0}

x2[t] = [-14/5]
[-1/5]
[0]
[1]

Answer 8

yep just like that xD

Answer 9

thank u :D! when you write the
x=[0]
y=[0]
z=[t]
solution, do you go according to row or column of the matrix ?

Answer 10

For any row of \(0\)s, I would assign to \(t,u,v,\ldots\)

Answer 11

So the first thing I did was say \(z=t\).

Answer 12

Now, the \(x\) row basically turned into: \[
1x+0y+0z = 0 \implies x=0
\]

Answer 13

ahhhh i understand! so next row would be: 0x + 1y + 0z = 0, y = 0

Answer 14

If the top row had been something like [1 0 2 | 4], then I would have gotten: \[
1x+0y+2z = 4\implies x = 4-2z \implies x=4-2t
\]

Answer 15

ah i see. gotcha! thanks a bunch mate :D !

Answer 16

Now, if the top row was something line [1 1 1 | 1]

Answer 17

There would be a \(y\) term, which would seem problematic.

Answer 18

But, I would have solved for \(y\) in the second column. So I could just substitute what I got for that in to the \(y\) term.

Answer 19

But yeah, do a more advanced problem and it will make more sense.

Answer 20

definitely! thank you so much :D

Answer 21

The only real trick to this is to convert from: \[
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&0
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}
\]to \[
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=
\begin{bmatrix}
0\\0\\t
\end{bmatrix}
\]Then it ends up being really simple.

Answer 22

OHHHH i see it

Answer 23

okay so for example
[1 1 0 | 0]
[0 0 1 | 0]
[0 0 0 | 0]

It would be
[-1]
t [0]
[1]

and if t = 1, then it would just be the same
[-1]
[0]
[1]!

Answer 24

Hold on...

Answer 25

Our matrix is: \[
\begin{bmatrix}
1&1&0\\
0&0&1\\
0&0&0
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}
\]We can swap rows:\[
\begin{bmatrix}
1&1&0\\
0&0&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}
\]Then we plug in the free variable:\[
\begin{bmatrix}
1&1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=
\begin{bmatrix}
0\\t\\0
\end{bmatrix}
\]Which gives us the equations:\[
x+y=0\implies x+t=0\implies x=-t\\
y=t\\
z=0
\]

Answer 26

\[
\begin{bmatrix}
x\\y\\z
\end{bmatrix}
=

\begin{bmatrix}
-1\\1\\0
\end{bmatrix}t
\]

Answer 27

oh crap, ah i see. you did the exact same thing you wrote above

Answer 28

I found it a bit confusing at first as well. They key thing to remember is that these are just systems of equations written in a weird way.

Answer 29

Want to see something interesting?

Answer 30

ahhh omg i see. and sure!

Answer 31

i'm doing eigenvectors and eigenvalues and it's just like wow i forgot how to do basic matrices lol

Answer 32

\[
\begin{bmatrix}
0&0\\
-m&1
\end{bmatrix}

\begin{bmatrix}
x\\y
\end{bmatrix} =
\begin{bmatrix}
0\\b
\end{bmatrix}
\]Suppose we just let the free variable for \(x\) be \(x\).\[

\begin{bmatrix}
1&0\\
-m&1
\end{bmatrix}

\begin{bmatrix}
x\\y
\end{bmatrix} =
\begin{bmatrix}
x\\b
\end{bmatrix}
\]This gives us equations: \[
x=x\\
-mx+y=b\implies y=mx+b
\]

Answer 33

Our line is just a system of equations, where \(x\) is a free variable.

Answer 34

i never thought of it like this with matrices omfg it's true

Answer 35

I think you can do the rest. The only tricks here are to swap your rows so that their first non-zero term is in the diagonal, then put in free variables.

Answer 36

thank you so much ^^

Answer 37

Or just recognize that if a column doesn't have a pivot, it's a free variable.


You're welcome.