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If T: $\Bbb R$3→ $\Bbb R$3 is a linear transformation such that:

$$ T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$ T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix} $$

Then the standard matrix for T is...

I'm not exactly sure how to approach this problem. Could anyone explain how to solve this problem?

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The standard matrix has columns that are the images of the vectors of the standard basis $$ T \Bigg (\begin{bmatrix}1\\0\\0\end{bmatrix} \Bigg), \qquad T \Bigg (\begin{bmatrix} 0\\1\\0 \end{bmatrix} \Bigg), \qquad T\Bigg (\begin{bmatrix}0\\0\\1 \end{bmatrix}\Bigg). \tag{1} $$ So one approach would be to solve a system of linear equations to write the vectors of the standard basis in terms of your vectors $$ \begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix}, \qquad \begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix}, \qquad \begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix}, $$ and then obtain (1).

Alternatively, note that if $A$ is the standard matrix you are looking for, then $$ A \cdot \begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix} = \begin{bmatrix} 5 & -4 & -6\\ 3 & 6 & -40 \\ 14 & -14 & -2 \\ \end{bmatrix}, $$ and multiply on the right by the inverse of $$ \begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix}. $$

Spoiler And the matrix $A$ is...

$$\begin{bmatrix}-1& 5& 3\\ 5& 3&-1\\-2& -2 & -4\end{bmatrix}$$

Many Thanks to @MartinSleziak for correcting two misprints in comments below.

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  • $\begingroup$ You should have $-2$ instead of $2$ in the position $2,2$ of the matrix for which you compute the inverse. WolframAlpha $\endgroup$ – Martin Sleziak Feb 25 '13 at 15:54
  • $\begingroup$ @MartinSleziak, many thanks, fixed now. $\endgroup$ – Andreas Caranti Feb 25 '13 at 16:09
  • $\begingroup$ @MartinSleziak, let me check, I suspect a typo on my side. $\endgroup$ – Andreas Caranti Feb 25 '13 at 16:23
  • $\begingroup$ @MartinSleziak, you were right, of course! $\endgroup$ – Andreas Caranti Feb 25 '13 at 16:27
  • $\begingroup$ @JyrkiLahtonen, thanks, and without my misprints, I'm sure! $\endgroup$ – Andreas Caranti Feb 25 '13 at 16:50
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You can put into a matrix given vectors and their images. If you then do elementary row operations, this property is not changed. (After each step you have in each row a vector and its image. This is because of linearity.) If you manage to obtain the identity matrix on the left, then you know the images of the vectors from the standard basis, which is sufficient to obtain the matrix of your linear transformation.

$ \left(\begin{array}{ccc|ccc} -2 & 3 & -4 & 5 & 3 & 14\\ 3 & -2 & 3 & -4 & 6 &-14\\ -4 & -5 & 5 & -6 &-40&-2 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 9 & 0\\ 3 & -2 & 3 & -4 & 6 &-14\\ -4 & -5 & 5 & -6 &-40&-2 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 9 & 0\\ 0 & -5 & 6 & -7 &-21 &-14\\ -4 & -5 & 5 & -6 &-40&-2 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 9 & 0\\ 0 & -5 & 6 & -7 &-21 &-14\\ 0 & -1 & 1 & -2 & -4 &-2 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 5 &-2\\ 0 & -5 & 6 & -7 &-21 &-14\\ 0 & -1 & 1 & -2 & -4 &-2 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 5 &-2\\ 0 & 1 & -1 & 2 & 4 & 2\\ 0 & -5 & 6 & -7 &-21 &-14 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 5 &-2\\ 0 & 1 & -1 & 2 & 4 & 2\\ 0 & 0 & 1 & 3 & -1 &-4 \end{array}\right)\sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 5 &-2\\ 0 & 1 & 0 & 5 & 3 &-2\\ 0 & 0 & 1 & 3 & -1 &-4 \end{array}\right) $

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    $\begingroup$ +1 A bunch of linear algebra students worldwide will appreciate this method. $\endgroup$ – Jyrki Lahtonen Feb 25 '13 at 16:28
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    $\begingroup$ like me so; +1. $\endgroup$ – mrs Feb 26 '13 at 2:59
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Note that if $$ \begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\begin{pmatrix}0 \\ 1\\ 0 \\ \end{pmatrix} -4\begin{pmatrix}0 \\ 0\\ 1 \\ \end{pmatrix} $$ so $$ T\begin{pmatrix}-2 \\ 3 \\ -4 \\ \end{pmatrix} = -2\times T\begin{pmatrix}1 \\ 0\\ 0 \\ \end{pmatrix} +3\times T\begin{pmatrix}0 \\ 1\\ 0 \\ \end{pmatrix} -4\times T\begin{pmatrix}0 \\ 0\\ 1 \\ \end{pmatrix} =-2T(\epsilon_1)+3T(\epsilon_2)-4T(\epsilon_3)$$ Now do the same for two other vectors to find out two relations written by to $T(\epsilon_i)$. Here, you have a system of 3 equations and 3 unknowns $T(\epsilon_i)$ which by solving that you get $T(\epsilon_i)_1^3$. Now use that fact that $$T\begin{pmatrix}x \\y \\ z \\ \end{pmatrix} = xT(\epsilon_1)+yT(\epsilon_2)+zT(\epsilon_3)$$ to find the original relation for $T$. I think by its rule you can find the associated matrix.

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