Another Linear Transformation

1. Jul 19, 2009

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

This is a slight variation of the last problem I posted.

Write the standard matrix representation for T1T2 and use it to find [T1T2(1,-3,0)]E.

2. Relevant equations

$$T_1\left(x_1,x_2,x_3\right)=\left(x_3,-x_1,x_3\right)$$

$$T_2\left(x_1,x_2,x_3\right)=\left(x_3-x_1,x_3-2x_2-x_1,x_1-x_3\right)$$

3. The attempt at a solution

$$T_1T_2=\left(x_3,-x_1,x_3\right)\cdot \left(x_3-x_1,x_3-2x_2-x_1,x_1-x_3\right)=x_1^2+2 x_1 x_2-x_1 x_3$$

$$A=\left(x_1^2+2 x_1 x_2-x_1 x_3\right)\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

Will A just end up being an identity matrix multiplied by the scalar that results from T1T2, or should I use a non-standard product for T1T2?

Last edited: Jul 19, 2009
2. Jul 19, 2009

Dick

You have it all wrong. T1*T2(x) means T1(T2(x)). That's nothing like T1(x)*T2(x) whatever that means. Find the matrices corresponding to T1 and T2 and multiply them.

3. Jul 19, 2009

Is this right, then?

$$A=\left( \begin{array}{ccc} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)\cdot \left( \begin{array}{ccc} -1 & 0 & 1 \\ -1 & -2 & 1 \\ 1 & 0 & -1 \end{array} \right)=\left( \begin{array}{ccc} 1 & 0 & -1 \\ 1 & 0 & -1 \\ 1 & 0 & -1 \end{array} \right)$$

4. Jul 19, 2009

Dick

Almost. But why are there two 1's in the third column of the first matrix?

5. Jul 19, 2009

Thanks!

There are two 1's in the third column of the first matrix because there is an x3 in the first and last element of T1 (is that the right terminology?)

6. Jul 19, 2009

Dick

You were right. My mistake.

7. Jul 19, 2009

Office_Shredder

Staff Emeritus
I made a mistake with that same transformation on another thread. There's something about that transformation... it's pretty sneaky

8. Jul 19, 2009

Dick

Must be us. DanielFaraday didn't have a problem.