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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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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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    $\begingroup$ 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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Let me propose an alternative way to solve this problem. There is a neat formula exactly for your case presented in "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely" $$ \vec{L}(\vec{p}) = (-1) \frac{ \det \begin{pmatrix} 0 & \vec{x} & \vec{y} & \vec{z} \\ p_1 & a_1 & b_1 & c_1 \\ p_2 & a_2 & b_2 & c_2 \\ p_3 & a_3 & b_3 & c_3 \\ \end{pmatrix} }{ \det \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{pmatrix} }, $$ where $\vec{L}$ is linear transformation acting on arbitrary point $\vec{p}$. $\vec{L}$ has the property $$ \vec{L}(\vec{a}) = \vec{x};\quad \vec{L}(\vec{b}) = \vec{y};\quad \vec{L}(\vec{c}) = \vec{z}. $$ Indices designate components of the corresponding vector.

Let's consider example. We need such $\vec{L}$ that $$ \vec{L}: \begin{pmatrix}-2\\ 3\\ -4\end{pmatrix} \mapsto \begin{pmatrix}5\\ 3\\ 14\end{pmatrix};~ \vec{L}: \begin{pmatrix}3\\ -2\\ 3\end{pmatrix} \mapsto \begin{pmatrix}-4\\ 6\\ -14\end{pmatrix};~ \vec{L}: \begin{pmatrix}-4\\ -5\\ 5\end{pmatrix} \mapsto \begin{pmatrix}-6\\ 40\\ -2\end{pmatrix}. $$ Now I plug them into the general expression $$ \vec{L}(\vec{p}) = (-1) \frac{ \det \begin{pmatrix} 0 & (5, 3, 14)^T & (-4, 6, -14)^T & (-6, 40, -2)^T \\ \begin{matrix} p_{1} \\ p_{2} \\ p_{3} \\ \end{matrix} & % \begin{matrix}-2\\ 3\\ -4\end{matrix} & % \begin{matrix}3\\ -2\\ 3\end{matrix} & % \begin{matrix}-4\\ -5\\ 5\end{matrix} \end{pmatrix} }{ \det \begin{pmatrix} \begin{matrix}-2\\ 3\\ -4\end{matrix} & % \begin{matrix}3\\ -2\\ 3\end{matrix} & % \begin{matrix}-4\\ -5\\ 5\end{matrix} \end{pmatrix} }. $$ Doing determinants I get $$ = \left[ 5\begin{pmatrix}5\\ 3\\ 14\end{pmatrix} + 5\begin{pmatrix}-4\\ 6\\ -14\end{pmatrix} + \begin{pmatrix}-6\\ 40\\ -2\end{pmatrix} \right] p_1 - \left[ 27\begin{pmatrix}5\\ 3\\ 14\end{pmatrix} + 26\begin{pmatrix}-4\\ 6\\ -14\end{pmatrix} + 6\begin{pmatrix}-6\\ 40\\ -2\end{pmatrix} \right] p_2 - \left[ 23\begin{pmatrix}5\\ 3\\ 14\end{pmatrix} + 22\begin{pmatrix}-4\\ 6\\ -14\end{pmatrix} + 5\begin{pmatrix}-6\\ 40\\ -2\end{pmatrix} \right] p_3 $$ or simplified $$ \vec{L}(\vec{p}) = \begin{pmatrix} -1 \\ 85 \\ -2 \end{pmatrix} p_1 + \begin{pmatrix} 5 \\ -477\\ -2 \end{pmatrix} p_2 + \begin{pmatrix} 3 \\ -401\\ -4 \end{pmatrix} p_3 . $$ Of course, you can write that in a vector form $$ \vec{L}(\vec{p}) = \begin{pmatrix} -1 & 5 & 3 \\ 85 & -477 & -401\\ -2 & -2 & -4 \end{pmatrix} \begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}. $$ Now you can easily check $$ \begin{pmatrix} -1 & 5 & 3 \\ 85 & -477 & -401\\ -2 & -2 & -4 \end{pmatrix} \begin{pmatrix}-2\\ 3\\ -4\end{pmatrix} = \begin{pmatrix}5\\ 3\\ 14\end{pmatrix};~ \begin{pmatrix} -1 & 5 & 3 \\ 85 & -477 & -401\\ -2 & -2 & -4 \end{pmatrix} \begin{pmatrix}3\\ -2\\ 3\end{pmatrix} = \begin{pmatrix}-4\\ 6\\ -14\end{pmatrix};~ \begin{pmatrix} -1 & 5 & 3 \\ 85 & -477 & -401\\ -2 & -2 & -4 \end{pmatrix} \begin{pmatrix}-4\\ -5\\ 5\end{pmatrix} = \begin{pmatrix}-6\\ 40\\ -2\end{pmatrix}; $$

For more details on the methods used, you can always refer to "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely". The latter contains many problems similar to this one as explained by the authors of the method presented.

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