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Let $S:R3→R3$ be the linear transformation $S(a,b,c)=(2b−2a, −b−2c,2a+c)$

Let $T:R3→R3$ be the linear transformation $T(a,b,c)=(a+b, −2b−2c, c−2a)$

Compute the composition $(S⋅T−T⋅S)(a,b,c)=... $

(Write your answer as a vector using a,b,c. For example $(3a+2b−3c,a+b))$

So, I tried to solve this question and according to formula first I need to find ($S⋅T$) then find ($T⋅S$) and at the end subtract them.

And I found $(S⋅T) = (-6b-4c-2a, 2b+2a, 2b+c)$ and $(T⋅S) =(b-2a-2c, 2b+2c-4a, 6a+c-4b)$

After I subtracted them I found $(-7b-2c, 6a-2c, 6b-6a)$ and when I submitted the answer it says that it's incorrect.

I checked my solutions a few times still didn't find any problems. I'm confused where did I do something wrong. Can someone help me with this?

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  • $\begingroup$ Looks perfectly fine to me. No errors. $\endgroup$ Apr 10, 2020 at 15:18
  • $\begingroup$ Check your calculation of $S\circ T$ again. The result is incorrect. $\endgroup$
    – amd
    Apr 11, 2020 at 2:28

1 Answer 1

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HINT

You can approach it alternatively through the matricial representation of the given operators.

Precisely speaking, let's consider the basis $\mathcal{B} = \{e_{1},e_{2},e_{3}\}$ for $\textbf{R}^{3}$. Thus we have \begin{align*} [S]_{\mathcal{B}} = [S(e_{1})^{T}, S(e_{2})^{T}, S(e_{3})^{T}] = \begin{bmatrix} -2 & 1 & 0\\ 0 & -1 & -2\\ 2 & 0 & 1 \end{bmatrix} \end{align*} Similarly, we do also have that \begin{align*} [T]_{\mathcal{B}} = [T(e_{1})^{T}, T(e_{2})^{T}, T(e_{3})^{T}] = \begin{bmatrix} 1 & 1 & 0\\ 0 & -2 & -2\\ -2 & 0 & 1 \end{bmatrix} \end{align*} Consequently, $[ST]_{\mathcal{B}} = [S]_{\mathcal{B}}[T]_{\mathcal{B}}$ and $[TS]_{\mathcal{B}} = [T]_{\mathcal{B}}[S]_{\mathcal{B}}$.

Can you take it from here?

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  • $\begingroup$ It seems like you solved with another method, that I don't know about. I just wanted to ask if the solution I found is correct or not, if it's ok with you, can you check it, please. $\endgroup$
    – User1331
    Apr 10, 2020 at 18:38
  • $\begingroup$ @Insomnia If you keep getting the wrong answer using a certain method, it’s a good idea to try a different one. $\endgroup$
    – amd
    Apr 11, 2020 at 2:38
  • $\begingroup$ I checked my calculations again but still didn't find any mistakes. If it's ok with you can you let me know the answer please, maybe I can see my mistakes from there $\endgroup$
    – User1331
    Apr 11, 2020 at 12:05

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