Let $\mathbf{M}$ be a matrix such that
\[\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{M} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}.\]Compute $\mathbf{M} \begin{pmatrix} 5 \\ 1 \end{pmatrix}.$
We can try solving for the matrix $\mathbf{M}.$  Alternatively, we can try to express $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$ as a linear combination of $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 5 \end{pmatrix}.$  Let
\[\begin{pmatrix} 5 \\ 1 \end{pmatrix} = a \begin{pmatrix} 2 \\ -1 \end{pmatrix} + b \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2a - 3b \\ -a + 5b \end{pmatrix}.\]Thus, $5 = 2a - 3b$ and $1 = -a + 5b.$  Solving, we find $a = 4$ and $b = 1,$ so
\[\begin{pmatrix} 5 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \begin{pmatrix} -3 \\ 5 \end{pmatrix}.\]Therefore,
\[\mathbf{M} \begin{pmatrix} 5 \\ 1 \end{pmatrix} = 4 \mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \mathbf{M} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = 4 \begin{pmatrix} 3 \\ 0 \end{pmatrix} + \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \boxed{\begin{pmatrix} 11 \\ -1 \end{pmatrix}}.\]
\begin{pmatrix}11\-1\end{pmatrix}
Level 3
Precalculus
Problem: Let $\mathbf{M}$ be a matrix such that
\[\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{M} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}.\]Compute $\mathbf{M} \begin{pmatrix} 5 \\ 1 \end{pmatrix}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the matrix $\mathbf{M}$, we need to express it in terms of its action on the given vectors. We have two equations involving $\mathbf{M}$:

1. $\mathbf{M} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}$
2. $\mathbf{M} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$

Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. We can set up a system of equations based on the matrix-vector multiplication.

From the first equation:
\[
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2a - b \\ 2c - d \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}
\]
This gives us two equations:
1. $2a - b = 3$
2. $2c - d = 0$

From the second equation:
\[
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} -3a + 5b \\ -3c + 5d \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}
\]
This gives us another two equations:
3. $-3a + 5b = -1$
4. $-3c + 5d = -1$

Now we have a system of four equations:
1. $2a - b = 3$
2. $2c - d = 0$
3. $-3a + 5b = -1$
4. $-3c + 5d = -1$

We can solve for $a$ and $b$ using equations 1 and 3. From equation 1, we can express $b$ in terms of $a$:
\[
b = 2a - 3
\]
Substituting this into equation 3:
\[
-3a + 5(2a - 3) = -1
\]
This simplifies to:
\[
-3a + 10a - 15 = -1 \implies 7a - 15 = -1 \implies 7a = 14 \implies a = 2
\]
Substituting $a = 2$ back into the expression for $b$:
\[
b = 2(2) - 3 = 4 - 3 = 1
\]

Next, we solve for $c$ and $d$ using equations 2 and 4. From equation 2, we express $d$ in terms of $c$:
\[
d = 2c
\]
Substituting this into equation 4:
\[
-3c + 5(2c) = -1
\]
This simplifies to:
\[
-3c + 10c = -1 \implies 7c = -1 \implies c = -\frac{1}{7}
\]
Substituting $c = -\frac{1}{7}$ back into the expression for $d$:
\[
d = 2\left(-\frac{1}{7}\right) = -\frac{2}{7}
\]

Now we have the matrix $\mathbf{M}$:
\[
\mathbf{M} = \begin{pmatrix} 2 & 1 \\ -\frac{1}{7} & -\frac{2}{7} \end{pmatrix}
\]

Next, we compute $\mathbf{M} \begin{pmatrix} 5 \\ 1 \end{pmatrix}$:
\[
\mathbf{M} \begin{pmatrix} 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -\frac{1}{7} & -\frac{2}{7} \end{pmatrix} \begin{pmatrix} 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2(5) + 1(1) \\ -\frac{1}{7}(5) - \frac{2}{7}(1) \end{pmatrix} = \begin{pmatrix} 10 + 1 \\ -\frac{5}{7} - \frac{2}{7} \end{pmatrix} = \begin{pmatrix} 11 \\ -1 \end{pmatrix}
\]

Thus, the final answer is $\begin{pmatrix} 11 \\ -1 \end{pmatrix}$.<|END_OF_LATENT|> The final answer is $\begin{pmatrix}11\-1\end{pmatrix}$.