Finding the Standard Matrix Representation for T1T2: A Non-Standard Product?

[DanielFaraday]

Homework Statement

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.

Homework Equations

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

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

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?

 

[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.

 

[DanielFaraday]

Is this right, then?

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

 

[Dick]

Is this right, then?

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

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

 

[DanielFaraday]

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?)

 

[Dick]

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?)

You were right. My mistake.

 

[Office_Shredder]

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

 

[Dick]

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