### Instruction: Find the $2 \times 2$ matrix $\mathbf{M}$ such that $\mathbf{M} \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 21 \end{pmatrix}$ and $\mathbf{M} \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -17 \end{pmatrix}.$.### Response: I notice that the two given vectors are linearly independent, which means that they form a basis for $\mathbb{R}^2$.
This implies that any other vector in $\mathbb{R}^2$ can be written as a linear combination of these two vectors.
Therefore, I can use them to construct a system of linear equations that relates the entries of $\mathbf{M}$ to the given transformations.
Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$
Then I can write: 
\[\mathbf{M} \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 21 \end{pmatrix} \implies 3a + 0b = 6 \text{ and } 3c + 0d = 21.\]
\[\mathbf{M} \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -17 \end{pmatrix} \implies -a + 5b = 3 \text{ and } -c + 5d = -17.\]
From the first two equations, I can easily solve for $a$ and $c$ by dividing by 3.
I get $a = 2$ and $c = 7$.
Substituting these values into the last two equations, I get a system of two equations in two unknowns: 
\[5b - 2 = 3 \text{ and } 5d - 7 = -17.\]
I can solve this system by adding 2 to both sides of the first equation and adding 7 to both sides of the second equation.
I get: 
\[5b = 5 \text{ and } 5d = -10.\]
Dividing by 5, I get $b = 1$ and $d = -2$.
Therefore, I have found all the entries of $\mathbf{M}$.