precalc

Using row properties of matrices:

\(\mathbf{A} \mathbf{u} = (2, 6, -2)\)

\(\mathbf{A} \mathbf{v} = (-2, 8, 12)\)

Hi!

In these type of problems, you should start by writing the givens.

That is, we are given u and v which are "three dimensional vectors", and so we can write:
Let 

\(u=\begin{pmatrix} u_1\\ u_2 \\ u_3 \end{pmatrix}\) , and \(v=\begin{pmatrix} v_1\\ v_2 \\ v_3 \end{pmatrix}\)

Now, we are given that the length of u and v is 2 and 4 respectively. Moreover, the angle between u and v is 120.

These givens should be familiar right? Yes: \(u*v = \left | u \right |\left | v\right |cos(\theta)\); this is the geometric definition of the dot product right?
Moreover, \(u*v=u_1v_1+u_2v_2+u_3v_3\) so we can calculate the dot product between u and v using the givens as follows:

\(u*v=u_1v_1+u_2v_2+u_3v_3=2(4)cos(120)=-4\) (Calculated from the geometric definition).

Now, how to continue this? Well, we are given a matrix's A rows. So why not construct this matrix?

So:

\(A:=\begin{bmatrix} u_1 && u_2 && u_3 \\ v_1 && v_2 && v_3 \\ 3u_1+2v_1 && 3u_2+2v_2 && 3u_3+2v_3 \end{bmatrix}\)

Now, what does the question wants? Au and Av right? Let's start with Au, and writing them yields:

\(A*u:=\begin{bmatrix} u_1 && u_2 && u_3 \\ v_1 && v_2 && v_3 \\ 3u_1+2v_1 && 3u_2+2v_2 && 3u_3+2v_3 \end{bmatrix} *\begin{pmatrix} u_1\\ u_2 \\ u_3 \end{pmatrix}\)

Well, this is a matrix multiplication (3*3 matrix with 3*1 matrix) so the resultant matrix will be (3*1) (A vector).

You may be used to multiply matrices that has numbers right? However, this matrices do not have numbers, rather variables.
But, it is the same procedure u_1 * u_1, u_2*u_2 etc.. , similarly for the second row, third row, and so we get:

   \(\implies Au=\begin{pmatrix} u_1^2+u_2^2+u_3^2\\ u_1v_1+u_2v_2+u_3v_3 \\ 3(u_1^2+u_2^2+u_3^2)+2(u_1v_1+u_2v_2+u_3v_3) \end{pmatrix}\)

For the last row, expand everything and group like terms to get what we got above.

(Multiply u by (3u+2v) etc.. and add all of these then factor the common factors).

But notice, the first row in Au is just the length of vector u squared! 

And, the second row is the dot product of u and v (which we found above to be -4)

And the third row is just a linear combination of both.

Hence,

\(Au=\begin{pmatrix} 2^2\\ -4\\ 3(2^2)+2(-4)\\ \end{pmatrix} = \begin{pmatrix} 4\\ -4 \\ 4 \end{pmatrix}\)

Now try to do the same steps for Av. 

I hope this helps, and if you need further help do not hesitate to ask!