Answer
See explanation
Work Step by Step
\[
T\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}-5 x_{2}+4 x_{3}, x_{2}-6 x_{3}\right)
\]
Goal
Show that $T$ is a linear transformation by finding a a matrix that implements the mapping.
Concepts
Definition of $A \mathbf{x}$
Solve
$\begin{aligned} T(\mathbf{x}) &=T\left(x_{1}, x_{2}, x_{3}\right) \\ &=\left[\begin{array}{c}x_{1}-5 x_{2}+4 x_{3} \\ x_{2}-6 x_{3}\end{array}\right] \\ &=\left[\begin{array}{c}x_{1} \\ 0 x_{1}\end{array}\right]+\left[\begin{array}{c}-5 x_{2} \\ x_{2}\end{array}\right]+\left[\begin{array}{c}4 x_{3} \\ 6 x_{3}\end{array}\right] \\=& x_{1}\left[\begin{array}{c}1 \\ 0\end{array}\right]+x_{2}\left[\begin{array}{c}-5 \\ 1\end{array}\right]+x_{3}\left[\begin{array}{c}4 \\ -6\end{array}\right] \\=&\left[\begin{array}{ccc}1 & -5 & 4 \\ 0 & 1 & -6\end{array}\right]\left[\begin{array}{c}x_{1} \\ x_{2}\end{array}\right] \\=& A(\mathbf{x}) \end{aligned}$
Conclusion
$A=\left[\begin{array}{ccc}1 & -5 & 4 \\ 0 & 1 & -6\end{array}\right]$
