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