Can someone help me with this very complex question? I don't even know where to start.

forsakenamth Jan 27, 2023

#1**0 **

Using row properties of matrices:

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

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

Guest Jan 27, 2023

#2**0 **

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!

Guest Jan 27, 2023

edited by
Guest
Jan 27, 2023

edited by Guest Jan 27, 2023

edited by Guest Jan 27, 2023