<Question>
A matrix $\mathbf{M}$ takes $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ to $\begin{pmatrix} 9 \\ 3 \end{pmatrix},$ and $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ to $\begin{pmatrix} 7 \\ -1 \end{pmatrix}.$  Find the image of the line $y = 2x + 1$ under $\mathbf{M}.$  Express your answer in the form "$y = mx + b$".
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the image of the line y = 2x + 1 under the matrix transformation M. The matrix M is given by its action on two vectors: it takes (2, -1) to (9, 3) and (1, -3) to (7, -1). Hmm, so first, I think I need to determine the matrix M itself. Once I have M, I can apply it to the line y = 2x + 1 and find the equation of the image line.

Wait, how do I find the matrix M? Since M is a linear transformation from R² to R², it can be represented as a 2x2 matrix. If I can find the columns of M, then I can write M explicitly. Let's recall that to find a matrix given its action on some vectors, those vectors need to form a basis. Are the given vectors (2, -1) and (1, -3) linearly independent? Let me check.

None

<prompt_1.1> Determine if the vectors (2, -1) and (1, -3) are linearly independent by calculating their determinant. </prompt_1.1>
-----

Alternatively, maybe another approach. Since linear transformations map lines to lines (or points if the line passes through the origin and the transformation is not invertible, but here the transformation is determined by two vectors which are linearly independent, so it's invertible).
Wait, are the two vectors (2,-1) and (1,-3) linearly independent?

<prompt_1.2> Check if the vectors (2, -1) and (1, -3) are linearly independent by calculating their determinant. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The determinant of the matrix formed by these vectors as columns is:

| 2  1 |
| -1 -3 |

Which is (2)(-3) - (1)(-1) = -6 + 1 = -5. Since the determinant is -5, which is not zero, the vectors are linearly independent and thus form a basis. Therefore, the matrix M can be determined using these vectors as a basis.

But wait, how exactly? Let me think. If I have a basis B = {v1, v2}, then the matrix [M]_B standard can be found by expressing M(v1) and M(v2) in terms of the standard basis. But here, we need to find M itself. Maybe I need to express M as [M]_E, where E is the standard basis. To do that, perhaps I can set up equations.

Let me denote M as:

[ a  b ]
[ c  d ]

Then, M*(2, -1) = (9, 3). Let's write that as:

2a - b = 9
2c - d = 3

Similarly, M*(1, -3) = (7, -1):

1a - 3b = 7
1c - 3d = -1
</execution_1.1>

<execution_1.2>
Let's compute the determinant of the matrix formed by them. The determinant is (2)(-3) - (-1)(1) = -6 +1 = -5 ≠ 0. So yes, they form a basis. Therefore, the matrix M is well-defined and invertible. That's good to know.

So, as before, M is:

[8, 7;
-1, -3]

Wait, let me just confirm once more. Let's verify that M*(2, -1) is (9,3). First component: 8*2 +7*(-1) = 16 -7 =9. Second component: -1*2 + (-3)*(-1) = -2 +3=1. Wait, that's (9,1), but according to the problem, it should be (9,3). Wait, that's a problem. There's a mistake here. Let me check my calculations again.

Wait, hold on. Earlier, when solving for a and b, I set up the equations correctly. Let me retrace. For the first basis vector (2,-1) mapping to (9,3):
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So now I have a system of two equations:

1. 2a - b = 9
2. 2c - d = 3

3. a - 3b = 7
4. c - 3d = -1

So, four equations with four variables: a, b, c, d. Let's solve them step by step.

<prompt_2.1> Solve the system of equations to find the values of a, b, c, and d for the matrix M. </prompt_2.1>
-----

Based on execution_1.2:
First, for the first component:

<prompt_2.2> Correct the mistake in determining the matrix M by solving the equations again. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Starting with the first two equations for a and b:

From equation 1: 2a - b = 9. Let's solve for b: b = 2a - 9.

Substitute this into equation 3: a - 3*(2a - 9) = 7.

Compute that:

a - 6a + 27 = 7

-5a + 27 = 7

-5a = 7 - 27

-5a = -20

a = (-20)/(-5) = 4.

So, a = 4. Then, substitute back into b = 2a -9: b = 2*4 -9 = 8 -9 = -1. So, b = -1.

Now for c and d. From equations 2 and 4:

From equation 2: 2c - d = 3. Let's solve for d: d = 2c -3.

Substitute into equation 4: c - 3*(2c -3) = -1.

Compute that:

c -6c +9 = -1

-5c +9 = -1

-5c = -1 -9

-5c = -10

c = (-10)/(-5) = 2.

So, c = 2. Then, substitute back into d = 2c -3: d = 2*2 -3 = 4 -3 = 1. So, d =1.

Therefore, the matrix M is:

[4  -1]
[2   1]

Let me check this with the given transformations.

First, apply M to (2, -1):

4*2 + (-1)*(-1) = 8 +1 =9

2*2 +1*(-1) =4 -1=3. That's correct, maps to (9,3).

Second, apply M to (1, -3):

4*1 + (-1)*(-3) =4 +3=7

2*1 +1*(-3)=2 -3=-1. Correct, maps to (7,-1).
</execution_2.1>

<execution_2.2>
2a + (-1)b = 9
So 2a - b =9

Second component:
2c + (-1)d =3
So 2c - d =3

Similarly, for the second basis vector (1,-3) mapping to (7,-1):

First component:
1a + (-3)b =7
So a -3b =7

Second component:
1c + (-3)d =-1
So c -3d =-1

So solving for a and b:

Equations:
1. 2a - b =9
2. a -3b =7

Let me solve equation 1 for b: b = 2a -9. Substitute into equation 2:

a -3*(2a -9) =7
a -6a +27 =7
-5a =7 -27
-5a = -20
a =4

Then, substitute a=4 into b=2a -9: b=8 -9= -1

So a=4, b=-1.

Similarly for c and d:

Equations:
3. 2c - d =3
4. c -3d =-1

Solve equation 3 for d: d=2c -3. Substitute into equation 4:

c -3*(2c -3) =-1
c -6c +9 =-1
-5c =-1 -9
-5c =-10
c=2

Then, substitute c=2 into d=2c -3: d=4 -3=1

Therefore, the matrix M is:

[4   -1]
[2    1]

Let me check again with the first vector:

M*(2, -1) = 4*2 + (-1)*(-1) =8 +1=9

Second component: 2*2 +1*(-1)=4 -1=3. Correct, that's (9,3).

For the second vector:

M*(1, -3) =4*1 + (-1)*(-3)=4 +3=7

Second component:2*1 +1*(-3)=2 -3=-1. Correct, that's (7,-1).
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, the matrix M is:
[4  -1]
[2   1]

Now, the problem is to find the image of the line y = 2x + 1 under this transformation. So, how do I find the image of a line under a linear transformation?

<prompt_3.1> Determine the approach to find the image of the line y = 2x + 1 under the linear transformation M. </prompt_3.1>
-----

Based on execution_2.2:
Perfect. So, matrix M is:
[4  -1]
[2   1]

Now, to find the image of the line y=2x+1 under this transformation.

<prompt_3.2> Reconfirm the matrix M by applying it to the given vectors again. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
I recall that the image of a line under a linear transformation is another line, provided the transformation is invertible. Since M is invertible (determinant is (4)(1) - (-1)(2) = 4 + 2 = 6 ≠ 0), the image should indeed be a line.

But how exactly to compute it? One approach is to parametrize the original line, apply the matrix transformation to the parametric equations, and then eliminate the parameter to find the equation of the image line. Alternatively, since the image of a line is a line, we can find two points on the original line, map them using M, and then find the equation of the line passing through the images. However, since the original line is infinite, we need to ensure that the image is correctly determined. But using two points might not be sufficient because the image line could be different if the original line is not in a special position (like passing through the origin or not). Wait, but any line can be parametrized, so maybe parametrizing the original line and then applying M would be the way to go.

Let me try that.
</execution_3.1>

<execution_3.2>
Let me proceed.

First, recall that the image of a line under a linear transformation is another line (assuming the transformation is invertible, which it is here since determinant of M is (4)(1) - (-1)(2) =4 +2=6 ≠0).

So, the image should be a line, and we need to find its equation in the form y = mx + b.

To find the image of the line y = 2x +1, we can parametrize the line and apply the transformation, then find the equation of the resulting line.

Alternatively, since the line is y=2x +1, we can write any point on the line as (x, 2x +1). Then, applying M to this point gives:

M*(x, 2x +1) = (4x + (-1)*(2x +1), 2x +1*(2x +1)) = (4x -2x -1, 2x +2x +1) = (2x -1, 4x +1)

So, the image point is (2x -1, 4x +1). Let's denote the image coordinates as (X, Y). So:

X = 2x -1

Y = 4x +1

We can solve for x in terms of X and substitute into the equation for Y to get Y in terms of X.

From X = 2x -1, solve for x:

X = 2x -1 → 2x = X +1 → x = (X +1)/2

Substitute into Y =4x +1:

Y =4*( (X +1)/2 ) +1 = 2*(X +1) +1 =2X +2 +1=2X +3

Therefore, Y =2X +3. So, the equation of the image line is y=2x +3.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
The original line is y = 2x + 1. Let me parametrize this line. Let x = t, so y = 2t + 1.

<prompt_4.1> Parametrize the line y = 2x + 1 and apply the matrix M to the parametric equations. </prompt_4.1>
-----

Based on execution_3.2:
Wait, that seems straightforward. Let me verify.

<prompt_4.2> Verify the result by checking the transformation of two points on the original line. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Therefore, any point on the line can be written as (t, 2t + 1), where t is a parameter. Now, apply the matrix M to this vector:

M*(t, 2t +1) = [4  -1; 2  1] * [t; 2t +1]

Compute the first component: 4*t + (-1)*(2t +1) = 4t -2t -1 = 2t -1

Second component: 2*t + 1*(2t +1) = 2t +2t +1 = 4t +1

So, the image under M of the point (t, 2t +1) is (2t -1, 4t +1). Now, we need to express this in terms of X and Y. Let me denote the image coordinates as (X, Y). So:

X = 2t -1

Y = 4t +1

Now, we have X and Y in terms of the parameter t. To eliminate the parameter t, we can solve one equation for t and substitute into the other.

From the first equation: X = 2t -1 → 2t = X +1 → t = (X +1)/2

Substitute this into the equation for Y:

Y = 4*( (X +1)/2 ) +1 = 2*(X +1) +1 = 2X +2 +1 = 2X +3

Therefore, the equation of the image line is Y = 2X +3. So, in the desired form, that's y = 2x +3.
</execution_4.1>

<execution_4.2>
Alternatively, let's test with two points on the original line. For example, take t=0: the point (0,1). Applying M:

M*(0,1) = (4*0 + (-1)*1, 2*0 +1*1) = (-1,1). So, the image point is (-1,1).

Take another point, say t=1: (1,3). Applying M:

M*(1,3) = (4*1 + (-1)*3, 2*1 +1*3) = (4 -3, 2 +3) = (1,5). So, the image points are (-1,1) and (1,5).

Now, let's see if these two points lie on the line Y =2X +3. For (-1,1):

2*(-1) +3 = -2 +3 =1. Correct. For (1,5):

2*1 +3 =5. Correct. So, the line passes through these two points. Therefore, the equation Y=2X +3 is correct.

But wait, how do we know that this is the entire image line and not some other line? Since the original line is parametrized by t, and the image is a parametrized curve. Since M is linear and the original curve is a line (affine), the image should also be a line. Therefore, two points determine the line, and since we have two points confirming the equation, the image is indeed Y=2X +3.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, is that correct? Let me check again.

<prompt_5.1> Verify the calculations for the matrix M and the parametrization of the line. </prompt_5.1>
-----

Based on execution_4.2:
Alternatively, another approach is to note that the image of the line can be found by considering the image of the direction vector of the original line and a particular point.

<prompt_5.2> Consider alternative methods to find the image of the line y = 2x + 1. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Original line: y=2x+1. Parametrization: x=t, y=2t+1. So, the vector is (t, 2t+1). Applying M:

First component: 4t + (-1)(2t +1) = 4t -2t -1 = 2t -1.

Second component: 2t +1*(2t +1) = 2t +2t +1 =4t +1.

So, yes, (X,Y)=(2t -1, 4t +1). Then solving for t from X: X =2t -1 → t=(X +1)/2.

Substitute into Y: Y=4*( (X +1)/2 ) +1 =2(X +1)+1=2X +2 +1=2X +3. So, Y=2X +3.
</execution_5.1>

<execution_5.2>
The original line y=2x +1 can be thought of as the point (0,1) (when x=0) plus t*(1,2), where t is a scalar. So, in vector form, any point on the line is (0,1) + t*(1,2). Then, applying the linear transformation M to this, we get:

M*(0,1) + t*M*(1,2)

We already computed M*(1,3) earlier, which is (1,5). Wait, but here we have M*(1,2). Wait, the direction vector of the original line is (1,2), since the slope is 2. So, we need to compute M*(1,2).

Wait, but we don't have M*(1,2) given. However, we can compute it. Let's do that.

Given that M is [4  -1; 2  1], then M*(1,2) = (4*1 + (-1)*2, 2*1 +1*2) = (4 -2, 2 +2) = (2,4). Therefore, the direction vector of the image line is (2,4), which is equivalent to (1,2). Then, the particular point on the original line is (0,1), which maps to M*(0,1) = (4*0 + (-1)*1, 2*0 +1*1) = (-1,1). Therefore, the image line passes through (-1,1) and has direction vector (1,2). Thus, the parametric equations of the image line are:

X = -1 + t*1

Y =1 + t*2

Then, eliminating t: from X = -1 + t → t = X +1. Substitute into Y: Y =1 +2*(X +1) =1 +2X +2 =2X +3. So, Y=2X +3. Same result as before.

Therefore, this confirms that the image line is Y=2X +3.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let me check if this is correct.

Wait, let me verify with another point. Let's take t=1 on the original line.

<prompt_6.1> Verify the equation of the image line by checking additional points. </prompt_6.1>
-----

Based on execution_5.2:
Alternatively, another approach: the image of the line can be found by expressing the original line in parametric form, apply the matrix, and then eliminate the parameter. Which is exactly what we did.

<prompt_6.2> Summarize the solution and confirm its correctness. </prompt_6.2>
</Plan_6>

<Execution_6>
<execution_6.1>
Then, the point is (1,3). Applying M gives (1,5). Let's see if (1,5) lies on Y=2X +3. 2*1 +3=5. Yes, correct. Another point: t= -1. Original point: (-1, y). y=2*(-1)+1= -2 +1= -1. So, point (-1, -1). Applying M:

M*(-1, -1) = (4*(-1) + (-1)*(-1), 2*(-1) +1*(-1)) = (-4 +1, -2 -1) = (-3, -3). Now, check if (-3, -3) lies on Y=2X +3. 2*(-3) +3= -6 +3= -3. Yes, correct. So, (-3, -3) is on the line Y=2X +3.
</execution_6.1>

<execution_6.2>
Alternatively, since the original line is affine, the image under a linear transformation is also affine, so it's a line. Therefore, we can also use the fact that the image of a line ax + by + c =0 under a linear transformation with matrix M is the set of points M(x,y) such that a x + b y + c =0. Wait, but in this case, we need to compute the image of the line y=2x +1. Let me see if there's another method.

Alternatively, write the line in parametric form, apply the transformation, and find the image. Which is what we did. So, seems consistent.

Alternatively, using the two-point method. Since the original line is y=2x +1, we can pick two points, map them, and find the equation of the line through their images. For example, take x=0 and x=1.

When x=0, y=1. So point (0,1). Image is (-1,1) as computed earlier.

When x=1, y=3. Point (1,3). Image is (1,5) as computed earlier.

So, the two points on the image line are (-1,1) and (1,5). Let's compute the equation of the line through these two points.

Compute the slope: (5 -1)/(1 - (-1)) =4/2=2. So slope is 2. Then, using point-slope form. Let's use (-1,1):

Y -1 =2(X - (-1)) → Y -1=2(X +1) → Y=2X +2 +1=2X +3. Same result. Perfect. So, this confirms again.

Thus, the image line is y=2x +3.
</execution_6.2>
</Execution_6>

<Plan_7>
Based on execution_6.2:
I think this is solid. Checked multiple ways: parametric, two points, direction vectors. All confirm the same result.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the image of the line $y = 2x + 1$ under the linear transformation $\mathbf{M}$ is the line $\boxed{y = 2x + 3}$.
</Final_answer>
</think>