# Homework Help: Linear Algebra double check?

1. Oct 4, 2011

### ElliottG

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I have gotten the matix multiplication uuT CORRECT! The only thing I can't get is the uTu part. I don't have an attempt at a solution because I have zero idea!

Second question:

1. The problem statement, all variables and given/known data

3. The attempt at a solution
Now, I have done this exactly per as in my notes (I hope?)

I applied the same row operations that are said in the question to the 3x3 identity matrix...yet it shows that some of them are wrong? ~85% of my answers are right but some of them aren't.

For instance, the "1/6" in E2 1st column 1st row is wrong (WTF?)

Double checks would be appreciated and even my methods!

Thanks,
Elliott

Last edited by a moderator: Apr 26, 2017
2. Oct 4, 2011

### HallsofIvy

1) Yes, your $uu^T$ is correct.
$$u^Tu= \begin{bmatrix}-4 & 2 & 7 \end{bmatrix}\begin{bmatrix}-4 \\ 2 \\ 7\end{bmatrix}$$
which is just like a "dot product" of the vector withitself.

2) Yes, (a) is correct. In (b) where did that "6" come from? An "elementary" matrix is, by definition, a matrix derived from the identity matrix by a single "row operation" and so can differ from the identity matrix in a single place. You are not combining (b) with (a) are you? They are completely separate questions.

Same thing in (c) and (d) you appear to be "accumulating" operations in each question and you are NOT asked to do that. Each answer should differ from the identity matrix in a single place.

Last edited by a moderator: Oct 4, 2011
3. Oct 4, 2011

### ElliottG

I see.

I don't understand your explanation for my question #1, though!

As for 2, I see what the problem is. I was accumulating operations because that's what we were doing in the notes for some reason...

Last edited by a moderator: Oct 4, 2011
4. Oct 4, 2011

### Staff: Mentor

uTu is the product of two matrices: a 1 x 3 matrix multiplying a 3 x 1 matrix. The product will be 1 x 1. For all intents and purposes, this is a scalar.

uTu produces the same value as u $\cdot$ u.