Okay, so I need to find the image of the line y = 2x + 1 under the matrix M. The matrix M is a 2x2 matrix because it's transforming vectors from R² to R². The problem gives me two specific vectors and their images under M. Let me start by recalling that to find a matrix given its action on two basis vectors, I can set up a system of equations. But here, it's not the standard basis; instead, we have two arbitrary vectors and their images. 

First, let me note the given vectors:

Original vectors:
v₁ = [2, -1]^T
v₂ = [1, -3]^T

Their images under M:
M(v₁) = [9, 3]^T
M(v₂) = [7, -1]^T

So, if I can express the standard basis vectors e₁ and e₂ in terms of v₁ and v₂, then I can find M(e₁) and M(e₂) by linearity. Once I have those, the matrix M can be formed as [M(e₁) | M(e₂)], and then I can apply it to the line y = 2x + 1.

Alternatively, since the image of a line under a linear transformation is another line (if the transformation is invertible), but since we don't know if M is invertible, maybe we need to check that first. Wait, but the problem says to express the answer as y = mx + b, so assuming that the image is a line. So, M must be invertible? Or maybe even if not, the image is still a line. Hmm.

But maybe the better approach is to parametrize the original line and then apply M to each point on the line. However, since the line is given by y = 2x + 1, parametrizing it might involve expressing it in terms of a parameter t, then applying M to each parameterized point, and then eliminating the parameter to find the equation of the image.

Alternatively, since M is a linear transformation, the image of the line can be found by transforming two points on the line and then finding the equation of the line through those two transformed points. But since the original line is y = 2x +1, which is an affine line, but linear transformations map affine lines to affine lines. So, the image of the line under M will also be a line. Therefore, if I can find two points on the original line, apply M to them, and then find the equation of the line through those two transformed points, that should give the answer.

But wait, the original line is y = 2x +1. To find two points on this line, I can choose two different x-values. Let's pick x = 0 and x = 1.

When x = 0, y = 1. So one point is (0, 1).

When x = 1, y = 3. So another point is (1, 3).

Therefore, the original line passes through (0,1) and (1,3). If I can find the images of these two points under M, then I can find the equation of the image line by finding the line through these two transformed points.

But wait, the problem is that we don't have M applied to (0,1) and (1,3). However, the given vectors are v₁ = [2, -1]^T and v₂ = [1, -3]^T, which are not (0,1) and (1,3). So, maybe we need to express (0,1) and (1,3) in terms of v₁ and v₂, then apply M.

Alternatively, since we have the action of M on v₁ and v₂, perhaps we can find the transformation matrix M, then apply it to the points (0,1) and (1,3).

Let me try that approach.

First, we need to find the matrix M. Since M is a 2x2 matrix, let's denote it as:

M = [a b]
    [c d]

Then, applying M to any vector [x; y] gives [a*x + b*y; c*x + d*y].

Given that M*v₁ = [9; 3], and M*v₂ = [7; -1], where v₁ = [2; -1] and v₂ = [1; -3].

So, writing out the equations:

For v₁:
a*2 + b*(-1) = 9
c*2 + d*(-1) = 3

For v₂:
a*1 + b*(-3) = 7
c*1 + d*(-3) = -1

So, we have four equations:

1) 2a - b = 9

2) 2c - d = 3

3) a - 3b = 7

4) c - 3d = -1

So, we can solve for a, b, c, d.

Starting with equations 1 and 3 for a and b:

Equation 1: 2a - b = 9

Equation 3: a - 3b = 7

Let me solve these two equations. Let's denote them as:

Equation 1: 2a - b = 9

Equation 3: a - 3b = 7

Let's solve equation 3 for a: a = 7 + 3b

Substitute into equation 1: 2*(7 + 3b) - b = 9

14 + 6b - b = 9

14 + 5b = 9

5b = 9 -14 = -5

b = -1

Then, a = 7 + 3*(-1) = 7 -3 = 4

So, a = 4, b = -1

Now, for c and d, using equations 2 and 4:

Equation 2: 2c - d = 3

Equation 4: c - 3d = -1

Let's solve equation 4 for c: c = -1 + 3d

Substitute into equation 2: 2*(-1 + 3d) - d = 3

-2 + 6d - d = 3

-2 + 5d = 3

5d = 5

d = 1

Then, c = -1 + 3*1 = 2

So, c = 2, d = 1

Therefore, the matrix M is:

[4  -1]
[2   1]

Let me verify this matrix with the given transformations.

First, M*v₁ = [4*2 + (-1)*(-1); 2*2 + 1*(-1)] = [8 +1; 4 -1] = [9; 3], which matches.

Second, M*v₂ = [4*1 + (-1)*(-3); 2*1 + 1*(-3)] = [4 +3; 2 -3] = [7; -1], which also matches. So, M is correct.

Great, so the transformation matrix M is:

[4  -1]
[2   1]

Now, the original line is y = 2x +1. I need to find the image of this line under M. 

One approach is to parametrize the original line and apply M to the parametrized points. Let's do that.

Parametrizing the line y = 2x +1. Let me let x = t, so y = 2t +1. Therefore, the parametrization is:

x = t

y = 2t +1

So, the parametric form is [t; 2t +1]. 

Now, applying M to this vector:

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

So, the image under M is the set of points (2t -1, 4t +1) for all real t. 

Now, to find the equation of the line in terms of y and x, we can eliminate the parameter t.

Let me denote the transformed coordinates as x' and y'. So,

x' = 2t -1

y' = 4t +1

We can solve for t from the first equation:

x' = 2t -1 => 2t = x' +1 => t = (x' +1)/2

Substitute into the second equation:

y' = 4*( (x' +1)/2 ) +1 = 2*(x' +1) +1 = 2x' +2 +1 = 2x' +3

Therefore, the equation is y' = 2x' +3, which can be written as y = 2x +3.

Wait, but let me check this. Let's verify with two points.

When t = 0:

Original point: (0, 1)

Image: [0*4 -1*(1); 0*2 +1*(1)] = [ -1; 1 ]

Wait, but according to the parametric equations above, when t=0:

x' = 2*0 -1 = -1

y' = 4*0 +1 = 1

So, the image point is (-1, 1). 

Using the equation y = 2x +3:

For x' = -1, y' should be 2*(-1) +3 = -2 +3 = 1. Which matches.

Another point, say t = 1:

Original point: (1, 3)

Image: [4*1 -1*3; 2*1 +1*3] = [4 -3; 2 +3] = [1; 5]

According to the parametric equations:

x' = 2*1 -1 = 1

y' = 4*1 +1 =5

Which matches. 

Now, applying the equation y = 2x +3 to x' =1: 2*1 +3 =5. Correct.

Another test: t = -1:

Original point: (-1, 2*(-1) +1) = (-1, -1)

Image under M: [4*(-1) -1*(-1); 2*(-1) +1*(-1)] = [-4 +1; -2 -1] = [-3; -3]

According to parametric equations:

x' = 2*(-1) -1 = -2 -1 = -3

y' = 4*(-1) +1 = -4 +1 = -3

Which matches. 

Testing the equation y = 2x +3 with x' = -3:

y' = 2*(-3) +3 = -6 +3 = -3. Correct.

So, all test points check out. Therefore, the image of the line y = 2x +1 under M is y = 2x +3.

Alternatively, another way to approach this is to note that the image of a line under a linear transformation is another line if the transformation is invertible. Since the determinant of M is (4)(1) - (-1)(2) = 4 +2 =6, which is not zero, so M is invertible. Therefore, the image of the line is indeed a line.

Alternatively, another method is to express the line in vector form and apply the transformation. The original line can be written as all vectors of the form (x, y) where y = 2x +1. So, (x, y) = (x, 2x +1). Then, applying M:

M*(x, 2x +1) = (4x -1*(2x +1), 2x +1*(2x +1)) = (4x -2x -1, 2x +2x +1) = (2x -1, 4x +1). Which is the same as before. Then, expressing this as a line in terms of x and y, eliminating x. So, x' = 2x -1 => x = (x' +1)/2. Then y' =4x +1 =4*( (x' +1)/2 ) +1 =2x' +2 +1=2x' +3. Therefore, y' =2x' +3. So, same result.

Therefore, the image of the line y =2x +1 under M is y =2x +3. 

Wait, but let me check once more. Suppose there's a different approach. For example, since M is linear, the image of the line is the set M(v) where v is in the original line. The original line is a set of points v = [x; 2x +1]. So, M(v) = [4x - (2x +1); 2x + (2x +1)] = [2x -1; 4x +1], which is the same as before. So, yeah, it's the same result.

Therefore, I think the answer is y =2x +3. All test points check out, the parametric equations are consistent, and the determinant is non-zero, so the transformation is invertible. Therefore, the image is a line, and the equation derived is correct.

**Final Answer**
The image of the line $y = 2x + 1$ under $\mathbf{M}$ is $\boxed{y = 2x + 3}$.
Therefore, the image of the line $y = 2x + 1$ under the matrix $\mathbf{M}$ is $\boxed{y = 2x + 3}$.
</think>