<Question>
Let $\mathbf{A}$ be a matrix such that
\[\mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} \quad \text{and} \quad \mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}.\]Find $\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix}.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the matrix A such that when it acts on the vectors (3,1,0) and (-5,2,2), it produces (2,4,-3) and (0,-5,-5) respectively. Then, using that matrix A, I have to compute its action on the vector (-13,3,4). Hmm, let me think step by step.

First, I remember that if a matrix is linear, which I assume it is because it's a matrix multiplication, then it can be determined if we know its action on a basis. The problem gives me two vectors that A acts on, but since A is a 3x3 matrix (because it's multiplying 3-dimensional vectors and giving 3-dimensional results), it needs three linearly independent vectors to uniquely determine the matrix.

<prompt_1.1> Determine if the given vectors form a basis and explain the implication for finding matrix A. </prompt_1.1>
-----

But wait, let me think again. The problem says "Let A be a matrix such that... ", so maybe A is given to exist, and we have to compute A times another vector. But since we can't determine A uniquely with only two vectors, perhaps the problem is constructed such that the third column can be inferred? Wait, but the two vectors (3,1,0) and (-5,2,2) are in R^3, and they might not span R^3. Let me check if they are linearly independent.

<prompt_1.2> Check if the given vectors span R^3 and discuss the implications for determining matrix A. </prompt_1.2>

But the problem gives us two equations. Let me write them out.

First equation: A * [3; 1; 0] = [2; 4; -3]

Second equation: A * [-5; 2; 2] = [0; -5; -5]

Let me denote matrix A as:

A = [a b c;
d e f;
g h i]

<prompt_1.3> Set up the system of equations from the given matrix-vector multiplications. </prompt_1.3>
-----

But let me verify this answer another way to make sure.

Alternatively, maybe there's another approach. Since we can't determine A uniquely, but maybe the vector (-13,3,4) can be expressed as a linear combination of the two given vectors, (3,1,0) and (-5,2,2). Let me check if that's possible.

<prompt_1.4> Try expressing the vector (-13,3,4) as a linear combination of the given vectors and verify the solution. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
So, maybe I need to check if these two vectors are linearly independent. Let me see.

Suppose c1*(3,1,0) + c2*(-5,2,2) = 0. Then, 3c1 -5c2 = 0, c1 + 2c2 = 0, and 0c1 + 2c2 = 0. From the third equation, 2c2 = 0 => c2 = 0. Then from the first equation, 3c1 = 0 => c1 = 0. So, they are linearly independent.
</execution_1.1>

<execution_1.2>
Calculating the determinant of the matrix formed by these vectors might help, but since it's a 3x3 matrix, two vectors can't span the entire space. Therefore, there are infinitely many matrices A that satisfy the given conditions. However, the problem is asking for A times a specific vector, (-13,3,4). Maybe this specific vector can be expressed as a linear combination of the two given vectors, allowing us to compute the result without needing the entire matrix?
</execution_1.2>

<execution_1.3>
Then, multiplying A by the first vector [3;1;0] gives:

First component: 3a + 1b + 0c = 2

Second component: 3d + 1e + 0f = 4

Third component: 3g + 1h + 0i = -3

Similarly, multiplying A by the second vector [-5;2;2] gives:

First component: -5a + 2b + 2c = 0

Second component: -5d + 2e + 2f = -5

Third component: -5g + 2h + 2i = -5

So, we have six equations:

1. 3a + b = 2

2. 3d + e = 4

3. 3g + h = -3

4. -5a + 2b + 2c = 0

5. -5d + 2e + 2f = -5

6. -5g + 2h + 2i = -5
</execution_1.3>

<execution_1.4>
So suppose (-13,3,4) = k1*(3,1,0) + k2*(-5,2,2). Let's solve for k1 and k2.

So:

3k1 -5k2 = -13

1k1 +2k2 = 3

0k1 +2k2 =4

Wait, from the third component: 0*k1 +2k2 =4 => 2k2=4 => k2=2.

Then from the second component: k1 +2*2=3 => k1=3-4= -1.

Then check the first component: 3*(-1) -5*2 = -3 -10= -13. Perfect, that works.

So (-13,3,4) = -1*(3,1,0) +2*(-5,2,2). Therefore, applying A to it:

A*(-13,3,4) = -1*A*(3,1,0) +2*A*(-5,2,2) = -1*(2,4,-3) +2*(0,-5,-5) = (-2, -4, 3) + (0, -10, -10) = (-2, -14, -7). Same result as before. So this confirms that the answer is correct.

Therefore, even though we can't determine A uniquely, because we don't have a third vector, but since the vector we need to apply A to can be expressed as a combination of the given vectors, we can compute the result without knowing the entire matrix A. That's a smarter way, using the linearity of matrix multiplication.

So both methods lead to the same answer, which is reassuring. Therefore, the final answer is (-2, -14, -7).
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Then, if we have a third vector that's linearly independent, we could set up equations to solve for the matrix entries. But since we don't have a third vector, maybe there's something missing here. Wait, the problem is to compute A times another vector, not to find the entire matrix A. Maybe we can express the vector (-13,3,4) as a linear combination of the two given vectors (3,1,0) and (-5,2,2). If we can do that, then we can use the linearity of matrix multiplication to compute A*(-13,3,4) without needing the entire matrix A.

Let me check if (-13,3,4) is in the span of the two vectors.

<prompt_2.1> Check if the vector (-13,3,4) is in the span of the given vectors and plan to use linearity to compute A*(-13,3,4). </prompt_2.1>
-----

Based on execution_1.3:
Now, we have six equations with nine variables, so we need to solve for these variables. Let me see if we can solve for some variables first.

<prompt_2.2> Solve the system of equations to express variables in terms of others and discuss the underdetermined system. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me set up the equation: a*(3,1,0) + b*(-5,2,2) = (-13,3,4). This gives us a system of equations:

3a - 5b = -13

a + 2b = 3

0a + 2b = 4
</execution_2.1>

<execution_2.2>
Looking at equations 1 and 4:

Equation 1: 3a + b = 2

Equation 4: -5a + 2b + 2c = 0

Similarly, equations 2 and 5:

Equation 2: 3d + e = 4

Equation 5: -5d + 2e + 2f = -5

Equations 3 and 6:

Equation 3: 3g + h = -3

Equation 6: -5g + 2h + 2i = -5
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let's solve the third equation first: 2b = 4 => b = 2.

<prompt_3.1> Solve the system of equations to find the values of a, b, and c. </prompt_3.1>
-----

Based on execution_2.2:
Let's solve equations 1 and 4 first.

<prompt_3.2> Solve the simplified system of equations to express variables in terms of others and identify the underdetermined system. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Then substitute b = 2 into the second equation: a + 4 = 3 => a = -1.

Now check the first equation: 3*(-1) -5*(2) = -3 -10 = -13. Correct. So, yes, (-13,3,4) is a linear combination of the first two vectors: -1*(3,1,0) + 2*(-5,2,2).
</execution_3.1>

<execution_3.2>
From equation 1: b = 2 - 3a.

Substitute into equation 4:

-5a + 2*(2 - 3a) + 2c = 0

-5a + 4 -6a + 2c = 0

-11a + 4 + 2c = 0

So, 2c = 11a - 4 => c = (11a -4)/2.

Similarly, equations 2 and 5:

From equation 2: e = 4 - 3d.

Substitute into equation 5:

-5d + 2*(4 - 3d) + 2f = -5

-5d + 8 -6d + 2f = -5

-11d + 8 + 2f = -5

2f = 11d -13 => f = (11d -13)/2.

Equations 3 and 6:

From equation 3: h = -3 -3g.

Substitute into equation 6:

-5g + 2*(-3 -3g) + 2i = -5

-5g -6 -6g + 2i = -5

-11g -6 + 2i = -5

2i = 11g +1 => i = (11g +1)/2.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, A*(-13,3,4) = A*(-1*(3,1,0) + 2*(-5,2,2)) = -1*A*(3,1,0) + 2*A*(-5,2,2) = -1*(2,4,-3) + 2*(0,-5,-5).
Let me compute that.

<prompt_4.1> Compute A*(-13,3,4) using the linearity of matrix multiplication. </prompt_4.1>
-----

Based on execution_3.2:
So, variables c, f, i are expressed in terms of a, d, g respectively. However, since we don't have any more equations, these variables can take any values depending on a, d, g. Therefore, matrix A is not uniquely determined. But wait, the problem is to compute A*(-13,3,4), and we already did that without needing the entire matrix. So even though A is not uniquely determined, the specific linear combination required for (-13,3,4) can still be computed because we expressed (-13,3,4) as a combination of the given vectors. Therefore, the answer is (-2, -14, -7).
-----

Based on execution_3.1:
Wait, but let me make sure. The problem might expect me to find matrix A explicitly, but since we can't determine it uniquely, maybe there's a trick here. Wait, but no, since the vector (-13,3,4) is in the span of the first two vectors, we can compute the result without knowing the entire matrix. So the answer is (-2, -14, -7).
-----

Therefore, the answer is \boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}.

<prompt_4.2> Finalize the answer and consider if matrix A is uniquely determined. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
First component: -1*2 + 2*0 = -2 + 0 = -2

Second component: -1*4 + 2*(-5) = -4 -10 = -14

Third component: -1*(-3) + 2*(-5) = 3 -10 = -7

So, the result is (-2, -14, -7). Wait, but let me verify this. Is this the correct approach? Since A is a linear transformation, if we can express the input vector as a linear combination of the given vectors, then applying A is just the same linear combination of the outputs. But for that, we need A to be linear, which it is, being a matrix multiplication. So yes, this approach is valid.
</execution_4.1>

<execution_4.2>
But let me think again. If the problem had given three linearly independent vectors and their images, then we could uniquely determine A. But here, with only two vectors, there's a whole line's worth of vectors (those not in their span) for which A is not determined. However, since the vector we need is in their span, the result is uniquely determined. So even though A isn't unique overall, the specific application to (-13,3,4) is.

So I think this is the correct approach and answer.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
**Final Answer**
\boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, $\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = \boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}$.
</Final_answer>
</think>