Question

How do I find out if a matrix is invertible or not by reducing it to a row-echelon form? As I am reducing the matrix, what should I look out for to determine if the matrix is invertible or not? Thanks.

Answer 1

if its determiniate = 0; its not invertible

Answer 2

Other than finding its determinant, say I wanna see if it is invertible by reducing it to a row-echelon form, what should I spot for?

Answer 3

If you don't end up with any rows/columns of zeros it is invertible. Every invertible matrix is equivalent (after elementary row operations) to the identity matrix.

Answer 4

I.E. If the matrix has linear dependencies in its rows/columns it is not invertible.

Answer 5

Thanks! Then say if I reduced a matrix until a row has 0, 0, 0. And I want the values of X1, x2 and x3, like this:

\[\begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 6\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}\]
Since the last row are all zeroes, is this linear system of equations going to have only one unique solution or infinitely many solutions?

Answer 6

Well how many free variables do you have?

Answer 7

Since there are x1, x2 and x3, can I say 3 free variables?

Answer 8

No. Only one variable is free. The number of free variables is the number of rows minus the number of pivot columns in your reduced form.

Answer 9

\[\begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 6\\ 0 & 0 & 0 \end{bmatrix} \equiv \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 0 & 0 \end{bmatrix} \implies x_2 = -2x_3, x_1 = -x_2-x_3\]

Answer 10

So \(x_3\) is your free variable.

Answer 11

and since you have at least one free variable there will be infinitely many solutions.

Answer 12

\[\begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 6\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \implies X \in Span\{\begin{bmatrix} 1\\-2 \\1\end{bmatrix}\}\]

Answer 13

How did you derive the part where you got \[x_{2}=-2x_{3} , x_{1}=-x_{2}-x_{3}\]?

Answer 14

\[\begin{bmatrix} 1 & 1 & 1\\ 0 & 3 & 6\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]\[\implies \left\{ \begin{array}{c} x_1 + x_2 + x_3 = 0\\ 3x_2 + 6x_3 = 0 \end{array}\right\}\]\[\implies \left\{ \begin{array}{c} x_1 = - x_2 - x_3\\ x_2 =- 2x_3 \end{array}\right\}\]

Answer 15

ohohoh.... I got that part on how you get x2=-2x3 and x1=-x2-x3 now. thanks!
But how do you determine if x3 is free variable one? Is it just because it is 0 throughout and I can just put any values to get the result of zero?

Also, I seem to only be able to find the combination of \[x_{1}=0 x_{2}=0 x_{3}=0\] to satisfy the whole equation. Like this is the only combination of values that I can get so that the 3 equations can equate to zero. Other than that, I can't get any other combination of values to equate to zero. How is this infinitely many solutions?

Answer 16

x3 is the free variable because we don't have an equation for it. I mean you can pick any one of them to be the free one, it doesn't matter. But typically you pick your free ones starting from the bottom.

There are infinitely many solutions because we don't have pivot columns for all of our variables. We have a free variable. So there are solutions to the system
\[\left\{ \begin{array}{c} x_1 + x_2 + x_3 = 0\\ 3x_2 + 6x_3 = 0 \end{array}\right\}\]
Namely that \(x_1 = x_3\) and \(x_2 = -2x_3\)
So you can pick any \(x_3\) you want. If you make \(x_2\) be -2 times the \(x_3\) you picked and \(x_1\) the same as the \(x_3\) you picked you will get \(\begin{bmatrix}0\\0\\0\end{bmatrix}\) when you multiply that X vector times your matrix.

Answer 17

oh...I think I am starting to see the picture now. Thank you so much for all the great explanation! Thank you so much! :D

Answer 18

Yeah, so that means that all the solutions can be described as:

\[X = \begin{bmatrix}x_3\\-2x_3\\x_3\end{bmatrix} = x_3\begin{bmatrix}1\\-2\\1\end{bmatrix}, \forall x_3 \in R \implies X \in Span\{\begin{bmatrix}1\\-2\\1\end{bmatrix}\}\]

Answer 19

What does the Span mean?

Answer 20

all linear combinations of the set of vectors

Answer 21

You haven't covered spans yet?

Answer 22

I have started on linear combinations but not aware that there is notation call Span.

So the Span means all combinations of the vector. From my understanding of linear combination, there is a multiplier. So this multiplier is just a single integer? Or is the multiplier also of a vector?

Answer 23

The multiplier is just some scalar constant.

\[Span\{\vec{v_1}, \vec{v_2}, ..., \vec{v_n}\} = c_1\vec{v_1} + c_2\vec{v_2} + ... + c_n\vec{v_n};\ \forall c_i \in R\]

Answer 24

Are the scalar constants any arbitary integers?

Answer 25

they can be any real number not just integers.

Answer 26

Ohh.. The scalar constant can be any integers or real numbers but must at least satisfy the system of equations, right? The scalar constant is just a number multiplied to a vector to kind of "extend" the vector in the same direction but longer length. So as long as any multiplier that can "extend" this vector and yet satisfy the system of equation, then that is a valid scalar constant. This is the right way to think about it?

Answer 27

The span doesn't satisfy any system. The span is just all the possible linear combinations. Every single possible real value for each of the constants times each of the vectors and added together will be an element of the span of those vectors.

In your problem we found that the solution to the system was a span of a single vector. So any real number times that vector will be a solution to your system.

Answer 28

I see... Thank you so much!! You've been very helpful! Thanks!! :D

Answer 29

Of course =)