Let $a,$ $b,$ $c$ be integers such that
\[\mathbf{A} = \frac{1}{5} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix}\]and $\mathbf{A}^2 = \mathbf{I}.$  Find the largest possible value of $a + b + c.$
We have that
\begin{align*}
\mathbf{A}^2 &= \frac{1}{25} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \\
&= \frac{1}{25} \begin{pmatrix} 9 + ab & -3a + ac \\ -3b + bc & ab + c^2 \end{pmatrix}.
\end{align*}Thus, $9 + ab = ab + c^2 = 25$ and $-3a + ac = -3b + bc = 0.$

From $9 + ab = ab + c^2 = 25,$ $ab = 16$ and $c^2 = 9,$ so $c = \pm 3.$

If $c = -3,$ then $-6a = -6b = 0,$ so $a = b = 0.$  But then $ab = 0,$ contradiction, so $c = 3.$  Thus, any values of $a,$ $b,$ and $c$ such that $ab = 16$ and $c = 3$ work.

We want to maximize $a + b + c = a + \frac{16}{a} + 3.$  Since $a$ is an integer, $a$ must divide 16.  We can then check that $a + \frac{16}{a} + 3$ is maximized when $a = 1$ or $a = 16,$ which gives a maximum value of $\boxed{20}.$
20
Level 5
Precalculus
Problem: Let $a,$ $b,$ $c$ be integers such that
\[\mathbf{A} = \frac{1}{5} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix}\]and $\mathbf{A}^2 = \mathbf{I}.$  Find the largest possible value of $a + b + c.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the integers \( a, b, c \) such that \( \mathbf{A}^2 = \mathbf{I} \), we start by calculating \( \mathbf{A}^2 \). 

First, we express \( \mathbf{A} \) as:
\[
\mathbf{A} = \frac{1}{5} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix}
\]

Next, we compute \( \mathbf{A}^2 \):
\[
\mathbf{A}^2 = \left( \frac{1}{5} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \right) \left( \frac{1}{5} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \right) = \frac{1}{25} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix}
\]

Calculating the product:
\[
\begin{pmatrix} -3 & a \\ b & c \end{pmatrix} \begin{pmatrix} -3 & a \\ b & c \end{pmatrix} = \begin{pmatrix} 9 + ab & -3a + ac \\ -3b + bc & ba + c^2 \end{pmatrix}
\]

Thus,
\[
\mathbf{A}^2 = \frac{1}{25} \begin{pmatrix} 9 + ab & -3a + ac \\ -3b + bc & ba + c^2 \end{pmatrix}
\]

Setting \( \mathbf{A}^2 = \mathbf{I} \) implies:
\[
\frac{1}{25} \begin{pmatrix} 9 + ab & -3a + ac \\ -3b + bc & ba + c^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]

This leads to the equations:
1. \( 9 + ab = 25 \)
2. \( -3a + ac = 0 \)
3. \( -3b + bc = 0 \)
4. \( ba + c^2 = 25 \)

From the first equation, we find:
\[
ab = 16 \quad \Rightarrow \quad b = \frac{16}{a} \quad (a \neq 0)
\]

From the second equation, factoring gives:
\[
a(c - 3) = 0 \quad \Rightarrow \quad a = 0 \text{ or } c = 3
\]

If \( a = 0 \), then from \( ab = 16 \), \( b \) must also be \( 0 \), which contradicts \( ab = 16 \). Thus, \( c = 3 \).

Substituting \( c = 3 \) into the third equation:
\[
-3b + 3b = 0 \quad \text{(always true)}
\]

Now substituting \( c = 3 \) into the fourth equation:
\[
ba + 9 = 25 \quad \Rightarrow \quad ba = 16
\]

We already have \( b = \frac{16}{a} \). Substituting this into \( ba = 16 \):
\[
\frac{16}{a} a = 16 \quad \text{(holds for any } a \neq 0\text{)}
\]

Now we maximize \( a + b + c \):
\[
a + b + c = a + \frac{16}{a} + 3
\]

Let \( f(a) = a + \frac{16}{a} + 3 \). To find the maximum, we differentiate:
\[
f'(a) = 1 - \frac{16}{a^2}
\]
Setting \( f'(a) = 0 \):
\[
1 = \frac{16}{a^2} \quad \Rightarrow \quad a^2 = 16 \quad \Rightarrow \quad a = 4 \text{ (since } a \text{ is positive)}
\]

Then, substituting \( a = 4 \):
\[
b = \frac{16}{4} = 4, \quad c = 3
\]
Thus,
\[
a + b + c = 4 + 4 + 3 = 11
\]

However, we need to check for other integer values of \( a \) that could yield a higher sum. Testing \( a = 8 \):
\[
b = \frac{16}{8} = 2, \quad c = 3
\]
Then,
\[
a + b + c = 8 + 2 + 3 = 13
\]

Testing \( a = 16 \):
\[
b = 1, \quad c = 3
\]
Then,
\[
a + b + c = 16 + 1 + 3 = 20
\]

Thus, the largest possible value of \( a + b + c \) is \( 20 \).<|END_OF_LATENT|> The final answer is $20$.