How to find $ker(A)$

Question

How can I find $ker(A)$ for

$A= \begin{pmatrix} 2 & 1 & 1 \\ 4 & 2 & 2 \\ 3 & 0 & 1 \end{pmatrix} $?

What does this represent?

Edit: OK, using the Hints i get

$2x+y+z=0$ $3x+z=0$

$y-x=0$ $y=x$

And now?

Answer 1

Straight from wikipedia:

$$\ker(A) = \left\{\mathbf{v}\in V : A\cdot \mathbf{x} = \mathbf{0}\right\}$$

In your case $V=\mathbb{R}^3$ so you want to find all vectors in the usual 3 space which make this true:

$$A\bf{x}= \begin{pmatrix} 2 & 1 & 1 \\ 4 & 2 & 2 \\ 3 & 0 & 1 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$

The science here is getting $A$ to reduced row echelon form. If that reduced matrix is $B$, then $\ker(A) = \ker(B)$.

Answer 2

So before I answer this we have to be clear with what objects we are working with here. Also, this is my first answer and I cant figure out how to actually insert any kind of equations, besides what I can type with my keyboard. sorry!

We have

ker(A)={x∈V:A⋅x=0}

This means if a vector x when applied to our system of equations (Matrix) are takin to the Zero Vector. This question is interested in finding all such vectors that do this.

Let A=[(2,1,1),(4,2,2),(3,0,1)] denote the matrix given at the top of the post.

Now lets find the vector x, such that Ax=0. Let x∈V, such that x=(x1,x2,x3)

Then Ax=[(2,1,1),(4,2,2),(3,0,1)]*(x1,x2,x3)

Which row reduces too

[(1,0,1/3),(0,0,0),(0,1,1/3)|(0,0,0)]

That is x1=(-1/3)x3, x2=(-1/3)x3

So, x=(x1,x2,x3)=((-1/3)x3,(-1/3)x3,x3)=x3(1,1,-3)

Now we have

ker(A)={x∈V:A⋅x=0}=Span{(1,1,-3)}

Note, Span{(1,1,-3)}=a(1,1,-3), for all real numbers a. That is the same thing as x3(1,1,-3), I just replaced the x3 with "a" since they both represent any real number.