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Given the linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^3$ where $$A = \begin{bmatrix} 1 & 2 & 1\\ 1 & -1 & 1\\ 2 & 1 & 1 \end{bmatrix}$$

Find $T(x)$ if

$$x =\begin{bmatrix} 2\\ 1\\ -4 \end{bmatrix}$$

What I did:

$T(x)$ = $Ax $ = $b$

So, I performed matrix multiplication $A \cdot b$ and got an answer

$$T(x) = \begin{bmatrix} 0\\ -3\\ 1 \end{bmatrix}$$

Can someone please verify this answer? I do not think this is the correct way to solve the problem.

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    The answer is correct. Why do you think it is not? – 2017-02-05
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    I think it is incorrect because A*b =/= b*A here (I know matrix multiplication is not commutative, but this would imply that there is more than one value of x.) – 2017-02-05
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    You are asked to find $b$ such that $T(x)=b$. Since $T(x)=Ax,$ the problem is to compute $Ax$ given $A$ and $x$. Here doesn't matter $xA,$ in fact, $xA$ is not defined in this case. – 2017-02-05
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    @stackofhay42 What do you mean by $bA$? This is not defined. – 2017-02-05
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    Oh I used an online calculator and it gave me an answer for $xA$. I guess this is correct then. Thanks! @littleO yes sorry I meant xA – 2017-02-05
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    @stackofhay42 $xA$ isn't defined either...you multiplication is correct, though. What is actually bothering you? – 2017-02-05

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