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HW1. Mathematical Foundations: Functions, Calculus, and Linear Algebra

Due date: 2026-01-22 23:59.
This homework is designed to help you assess and refresh some of the background knowledge that will be important for the rest of the course. Completing this assignment will help identify any areas that may need review and set you up for success in the rest of the course.

Functions: Preimage Calculation in 1D

For the following problems, consider the definition of the preimage of a real-valued function in one dimension:
To visualize preimages in 1D, you will use the following program:
PreimageCalculator1D
After running the program, you can select any function f(x) that maps real numbers to real numbers, and any value y and closed interval domain X = [x1, x2] to visualize preimages.

How to reason about preimages

The preimage of a value 𝑦 under a function 𝑓 is the set of all inputs π‘₯ in the domain such that 𝑓(π‘₯) = 𝑦.
A reliable way to check whether a particular point π‘₯ belongs to the preimage of 𝑦 is to substitute π‘₯ into the function and evaluate 𝑓(π‘₯).
Example:
f(x) = (x - 5)^3 + 1. \\\
To check whether π‘₯ = 6 belongs to the preimage of 𝑓(π‘₯) = 2, compute:
f(6) = (6 - 5)^3 + 1 = 1 + 1 = 2. \\\
Since the result equals 𝑦, the point π‘₯ = 6 belongs to the preimage of 𝑓(π‘₯) = 2.

⚠️ Note About the Visualizer

No visualizer is perfect. Computing roots of real functions is prone to numerical errors and sampling limitations. As a result, the visualizer may:
These limitations are common in numerical analysis. This is why it is equally important to learn how to reason about preimages analytically.

Question 1

When the code runs for f(x) = x**3 - 3*x**2 + 2 and y = 1, how many preimage elements (marked as green dots) does the visualization show in the interval [-1, 3]?
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Question 2

Consider the following functions, domains, and values. For which of these combinations does the visualizer report an empty preimage?
A. f(x) = x**2 - 2, X = [-3, 3], y = 2
B. f(x) = fabs(x - 1), X = [-2, 5], y = 2
C. f(x) = sin(pi*x) + 0.5, X = [0, 4], y = 1.5
D. f(x) = (x - 1)**4 + 2, X = [-1, 3], y = 2
E. f(x) = exp(-x) - 0.2, X = [0, 5], y = -0.1
F. f(x) = sin(pi*x) + 0.5, X = [0, 4], y = 0.5
G. f(x) = (x - 1)**3 + x%2, X = [0, 3], y = 3
H. None of the above.
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Question 3

Among the function, domain, and observation combinations listed in Question 2, identify a case for which the visualizer did not compute the preimage correctly - that is, a case in which the mathematically correct preimage differs from what the visualizer reports.
A. f(x) = x**2 - 2, X = [-3, 3], y = 2
B. f(x) = fabs(x - 1), X = [-2, 5], y = 2
C. f(x) = sin(pi*x) + 0.5, X = [0, 4], y = 1.5
D. f(x) = (x - 1)**4 + 2, X = [-1, 3], y = 2
E. f(x) = exp(-x) - 0.2, X = [0, 5], y = -0.1
F. f(x) = sin(pi*x) + 0.5, X = [0, 4], y = 0.5
G. f(x) = (x - 1)**3 + x%2, X = [0, 3], y = 3
H. None of the above.
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Question 4

Consider the following function, domain, and value: f(x) = x%2 + fabs(x) + int(x), y = 1, X=[-2, 2]. Which of the following is the correct preimage?
A. (-2, -1) U {0} U {0.5}
B. (-2, -1) U {0.5}
C. (-2, -1) U {0}
D. (-2, -1)
E. The preimage is an empty set.
F. None of the above.
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Calculus: Single-Variable Integration

For the next set of problems, we will consider a point moving along the real line. We will examine how the point’s position, velocity, and acceleration change over time. We will use the following definitions:
To visualize the relationship between acceleration, velocity, and position, we will use the following program:
Integration1D
Initially, the program is set up with the following parameters. From time t = 2 to t = 3, the acceleration of a point on the line is ẍ = 2. From time t = 3 to t = 4, its acceleration is ẍ = βˆ’3. At all other times, the acceleration is zero.
The acceleration is double integrated over the time interval defined by tβ‚€ = 0, t₁ = 10. Thus, the acceleration can be expressed as ẍ(t) = (2 <= t)(t < 3)2 + (3 <= t)(t < 4)(-3) over this interval of time.
The initial conditions are set as follows: at time tβ‚€ = 0, the point is at position x(0) = 0, and its velocity is αΊ‹(0) = 1. The program then numerically computes the values of position and velocity at the end of the time interval, x(10) and αΊ‹(10), by integrating acceleration values.
We will practice visualizing the integration process by running the program. You can define any acceleration function ẍ(t), integration interval [tβ‚€, t₁], and initial conditions x(tβ‚€) and αΊ‹(tβ‚€) to observe how the corresponding velocity αΊ‹(t) and position x(t) evolve over time. We will also practice how these results can be computed analytically.

From Code to Mathematical Representation

In the visualizer, the acceleration of the point is specified using Python-style indicator expressions, such as
(2<=t)*(t<3)*2. The boolean expressions (2<=t) and (t<3) evaluate to either 0 or 1, depending on the value of 𝑑, which creates a compact way to describe piecewise-defined functions.
The same acceleration can be written mathematically as a piecewise function:
\ddot{x}(t)= \begin{cases} 2, & 2 \le t < 3, \\ -3, & 3 \le t < 4, \\ 0, & \text{otherwise}. \end{cases} \\\
Together with the initial conditions
\dot{x}(0)=1, \qquad x(0)=0, \\\
and the time interval
t \in [0,10], \\\
this fully specifies the motion of the point along the line for this time interval.
The code-based representation used by the visualizer and the mathematical piecewise definition above describe exactly the same physical behavior. They are simply two different ways to describe the same motion of a point along the line.

Relationship between acceleration, velocity, and position

Finally, recall the fundamental relationships between acceleration, velocity, and position for motion along a line.
Velocity is obtained by accumulating acceleration over time. If the velocity at time tβ‚€ is known, then for any later time t it can be written as
\dot{x}(t) = \dot{x}(t_0) + \int_{t_0}^{t} \ddot{x}(s)ds. \\\
Similarly, position is obtained by accumulating velocity over time. If the position at time tβ‚€ is known, then
x(t) = x(t_0) + \int_{t_0}^{t} \dot{x}(s)ds. \\\
These expressions formalize the idea that acceleration determines how velocity changes, and velocity determines how position changes.
In the visualizer, these integrations are performed numerically. In the questions that follow, you will practice expressing the same relationships analytically by writing down the corresponding integrals over the time intervals where the acceleration is nonzero.

Question 5

Run the visualization program and use the default settings:
ẍ(t) = (2<=t)*(t<3)*2 + (3<=t)*(t<4)*(-3)
αΊ‹(tβ‚€) = 1
x(tβ‚€) = 0
Domain [tβ‚€, t₁] = [0, 10]
Now, you can click Compute & Plot to generate the acceleration, velocity, and position plots for the moving point. You can also click Play/Pause/Reset buttons to observe the point move.
Which of the following correctly describes the velocity of the point?
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Question 6

Consider a slightly different setup:
\begin{aligned} &\ddot{x}(t) = \begin{cases} 2, & 1 \le t < 3,\\[4pt] -3, & 3 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[8pt] &\dot{x}(0) = 0,\\[4pt] &x(0) = 0,\\[4pt] &t \in [0,\,10]. \end{aligned} \\\
Which of the following correctly expresses the velocity of the point at time t = 10?
\begin{aligned} \mathsf{A.}\;& \dot{x}(10) = \int_{1}^{3} 2\,ds\\ \\ \mathsf{B.}\;& \dot{x}(10) = \int_{1}^{3} 2\,ds + \int_{3}^{4} (-3)\,ds\\ \\ \mathsf{C.}\;& \dot{x}(10) = \int_{0}^{10} (-3)\,ds\\ \\ \mathsf{D.}\;& \dot{x}(10) = \int_{1}^{3} 2\,ds - \int_{3}^{4} 3\,ds - \int_{4}^{10} 1\,ds \\ \\ \mathsf{E.}\;& \dot{x}(10) = 0\\ \\ \mathsf{F.}\;& \text{None of the above.}\\ \end{aligned}
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Question 7

Now consider another setup:
\begin{aligned} &\ddot{x}(t) = \begin{cases} t, & 2 \le t < 3,\\[4pt] t-6, & 3 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[8pt] &\dot{x}(0) = 1,\\[4pt] &x(0) = 0,\\[4pt] &t \in [0,\,10]. \end{aligned}
Which of the following correctly expresses the velocity of the point on the interval t=[2,3)?
\begin{aligned} \mathsf{A.}\;& \dot{x}(t) = 1 + \int_{2}^{t} s\,ds = 1+ \tfrac{1}{2}(t^2 - 4) \\[8pt] \mathsf{B.}\;& \dot{x}(t) = 1 + \int_{0}^{t} s\,ds = 1 + \tfrac{1}{2}t^2 \\[8pt] \mathsf{C.}\;& \dot{x}(t) = t^2 - 4t + 4 \\[8pt] \mathsf{D.}\;& \dot{x}(t) = \tfrac{1}{2}(t - 2)^2 \\[8pt] \mathsf{E.}\;& \dot{x}(t) = \int_{2}^{t} (s - 6)\,ds = \tfrac{1}{2}(t^2 - 4t - 8) \\[8pt] \mathsf{F.}\;& \text{None of the above.} \end{aligned}
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Question 8

Find one setup among the following in which the point moves in such a way that it returns to the same position where it started at tβ‚€ = 0 and comes to a complete halt at that point.
\begin{aligned} \mathsf{A.}\;& \ddot{x}(t) = \begin{cases} 2, & 0 \le t < 2,\\[4pt] -2, & 2 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[6pt] &\dot{x}(0) = 0, \quad x(0) = 0, \quad t \in [0,\,6]. \\[12pt] \mathsf{B.}\;& \ddot{x}(t) = \begin{cases} 1, & 0 \le t < 1,\\[4pt] -3, & 1 \le t < 2,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[6pt] &\dot{x}(0) = 0, \quad x(0) = 0, \quad t \in [0,\,3]. \\[12pt] \mathsf{C.}\;& \ddot{x}(t) = \begin{cases} -2, & 0 \le t < 2,\\[4pt] 2, & 2 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[6pt] &\dot{x}(0) = 1, \quad x(0) = 0, \quad t \in [0,\,6]. \\[12pt] \mathsf{D.}\;& \ddot{x}(t) = \begin{cases} 3, & 0 \le t < 2,\\[4pt] -3, & 2 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[6pt] &\dot{x}(0) = 0, \quad x(0) = 0, \quad t \in [0,\,6]. \\[12pt] \mathsf{E.}\;& \ddot{x}(t) = \begin{cases} t - 3, & 2 \le t < 4,\\[4pt] 0, & \text{otherwise,} \end{cases}\\[6pt] &\dot{x}(0) = 0, \quad x(0) = 0, \quad t \in [0,\,6]. \\[12pt] \mathsf{F.}\;& \text{None of the above.} \end{aligned}
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Linear Algebra: Singular Value Decomposition (SVD)

For any real-valued π‘š Γ— 𝑛 matrix, 𝑀, the singular value decomposition (SVD) allows us to write:
M = U \, \Sigma \, V^{\mathsf{T}}, \\\
in which,
In this part of the assignment, we will explore how SVD decomposes general 3Γ—2 matrices into 𝑀 = π‘ˆ Ξ£ 𝑉ᡀ, and how this decomposition helps us:

About Matrix Sizes

In this homework, 𝑀 is a 3 Γ— 2 matrix. It maps 2D inputs to 3D outputs. As a result:
In general, SVD works for any real matrix, not just 3 Γ— 2 ones.

Visualization Tool

You will use the following visualization tool:
SVDPlayer3D
This tool visualizes how a 2D unit circle in the input (π‘Ž, 𝑏)-plane is mapped to a 3D ellipse by the matrix 𝑀. The program shows this transformation in two different ways:
The visualizer animates these steps, so you can clearly observe the effect of each matrix independently.
Note: Just like other visual tools in this homework, this visualizer is also subject to numerical limitations.
Understanding the visualization is important, but it is equally important to be able to reason about matrix behavior analytically.

Example

Consider the matrix:
M= \begin{bmatrix} 2 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix}. \\\
This matrix takes a 2D point (π‘₯, 𝑦) and maps it into 3D by stretching the second coordinate more than the first, and embedding the result in a 3D space:
M \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 2 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 2x \\ y \\ 0 \end{bmatrix}. \\\
One of its singular value decomposition is:
M = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \\\
in which:
\Sigma= \begin{bmatrix} 2 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix}, \\\
In this simple case, it is trivial to see how SVD separates identity rotations from scaling (singular values). More complicated matrices behave the same way, except that rotations become more complicated.

Question 9

In this question, you will explore the geometric meaning of the singular value decomposition, 𝑀 = π‘ˆ Ξ£ 𝑉ᡀ.
Run the SVDPlayer3D program. Start with any matrix 𝑀 and switch to SVD Steps Mode. Watch the animation carefully and pay attention to what each matrix does to answer the following conceptual question. Use the button Random 𝑀 (well-cond.) in the program to produce a random well conditioned matrix if needed.
Which of the following correctly describes the geometric meaning of each SVD component?
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Question 10

Run the program again in SVD Steps Mode.
Use a matrix 𝑀 where the SVD produces clearly identifiable transformations (use the button Random 𝑀 (well-cond.) in the program if needed).
Watch the animation again. Which of the following best describes the difference between π‘ˆ and 𝑉?
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Question 11

Consider the matrix
M = \begin{bmatrix} 0 & 1\\ 1 & 1\\ 1 & 0 \end{bmatrix}
and consider its singular value decomposition:
U = \begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix}, \qquad \Sigma = \begin{bmatrix} \sqrt{3} & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}, \qquad V^{\!\top} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}.
Which is the correct expression for the pseudoinverse of 𝑀? (You do not need to multiply them; just choose the correct construction.)
\begin{aligned} \mathsf{A.}\;& M^{+} = \begin{bmatrix} \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & -\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \begin{bmatrix} \sqrt{3} & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \\[14pt] \mathsf{B.}\;& M^{+} = \begin{bmatrix} \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & -\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \begin{bmatrix} \frac{1}{ \sqrt{3}} & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \\[14pt] \mathsf{C.}\;& M^{+} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \sqrt{3} & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \\[14pt] \mathsf{D.}\;& M^{+} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{ \sqrt{3}} & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & 0 & -\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \\[14pt] \mathsf{E.}\;& M^{+} = \begin{bmatrix} \frac{1}{ \sqrt{3}} & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \\[14pt] \mathsf{F.}\;& \text{None of the above.} \end{aligned}
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Question 12

Which matrix will produce no rotation in the π‘ˆ-step (that is, the SVD of 𝑀 computed by the visualizer has π‘ˆ equal to the identity matrix)?
\begin{aligned} \mathsf{A.}\;& M = \begin{bmatrix} 2 & 0\\ 0 & 1\\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{B.}\;& M = \begin{bmatrix} 0 & 2\\ 1 & 0\\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{C.}\;& M = \begin{bmatrix} 1 & 1\\ 0 & 1\\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{D.}\;& M = \begin{bmatrix} 1 & 0\\ 1 & 1\\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{E.}\;& M = \begin{bmatrix} 1 & 0\\ 0 & 1\\ 1 & 0 \end{bmatrix} \\[12pt] \mathsf{F.}\;& \text{More than one above.}\\ \mathsf{G.}\;& \text{None of the above.} \end{aligned}
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Question 13

Which matrix will make the SVD visualizer produce a Ξ£ with equal singular values, so a circle is only uniformly scaled?
\begin{aligned} \mathsf{A.}\;& M = \begin{bmatrix} 1 & \sqrt{3} \\ \sqrt{3} & -1 \\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{B.}\;& M = \begin{bmatrix} \sqrt{2} & 1 \\ 0 & \sqrt{2} \\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{C.}\;& M = \begin{bmatrix} 1 & \sqrt{2} \\ \sqrt{2} & 1 \\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{D.}\;& M = \begin{bmatrix} \sqrt{3} & 0 \\ 1 & 2 \\ 0 & 0 \end{bmatrix} \\[12pt] \mathsf{E.}\;& M = \begin{bmatrix} \sqrt{5} & 1 \\ 1 & \sqrt{5} \\ 0 & 0 \end{bmatrix} \\[14pt] \mathsf{F.}\;&\text{More than one above.}\\ \mathsf{G.}\;& \text{None of the above.} \end{aligned}
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Question 14

Which matrix keeps the transformed ellipse entirely in the original plane?
\begin{aligned} \mathsf{A.}\;& M= \begin{bmatrix} 2 & 0\\ 0 & 1\\ 1 & 1 \end{bmatrix} \\[14pt] \mathsf{B.}\;& M= \begin{bmatrix} 2 & 1\\ 0 & 1\\ 0 & 1 \end{bmatrix} \\[14pt] \mathsf{C.}\;& M = \begin{bmatrix} 1 & \sqrt{2} \\ \sqrt{2} & 1 \\ 0 & 0 \end{bmatrix} \\[14pt] \mathsf{D.}\;& M= \begin{bmatrix} 2 & 0\\ 0 & 1\\ 1 & 0 \end{bmatrix} \\[14pt] \mathsf{E.}\;& M = \begin{bmatrix} 1 & \sqrt{3} \\ \sqrt{3} & -1 \\ 0 & 0 \end{bmatrix} \\[14pt] \mathsf{F.}\;& \text{More than one above.}\\ \mathsf{G.}\;& \text{None of the above.} \end{aligned}
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Question 15

Which of the following matrices 𝑀 relate the two points through the transformation:
𝑝_1=𝑀^{+}𝑝_2, \ p_1 = \begin{bmatrix} 0 \\ 3 \end{bmatrix}, p_2 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}. \\\
Hint: You may find the following relation useful:
𝑀^{+} = (M^{\top} M)^{-1} M^{\top} \\\
and, therefore:
M^{\top} M 𝑝_1= M^{\top} 𝑝_2. \\\
\begin{aligned} \mathsf{A.}\;& M = \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ -2 & 0 \end{bmatrix} \\[12pt] \mathsf{B.}\;& M = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ -3 & 1 \end{bmatrix} \\[12pt] \text{C.}\;& M = \begin{bmatrix} -\frac{1}{3} & 1 \\ 0 & 0 \\ -\frac{1}{3} & 1 \end{bmatrix} \\[12pt] \mathsf{D.}\;& M = \begin{bmatrix} \sqrt{3} & -1 \\ 0 & 1 \\ -3 & 0 \end{bmatrix} \\[12pt] \mathsf{E.}\;& M = \begin{bmatrix} -1 & 0 \\ -1 & 1 \\ -1 & 2 \end{bmatrix}\\[12pt] \mathsf{F.}\;&\text{More than one above.}\\ \mathsf{G.}\;& \text{None of the above.} \end{aligned}
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Authors

Anna LaValle.
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