Final Review I

Lecture 38

Author
Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

April 20, 2026

Overview

Final 1 Information

  • Final 1 will be held during class on Friday, April 24, 2026 from 11:00 AM to 11:50 AM.
  • Please bring an electronic device that can access WebAssign.
  • Be logged in and ready before 11:00 AM to avoid technical issues.
  • The exam is closed book. No materials are allowed, including the course website.
    • You may use blank scratch paper and a pen, or a tablet/iPad for writing.
    • If you use a tablet, only a blank writing app (white page) may be open. No other apps or materials may be open.
  • A traditional calculator is allowed, though you will likely not need it.

Final 2 Information

  • Final 2 will be held on Monday, April 27, 2026, from 10:30 AM to 12:30 PM, following the registrar’s schedule.
  • This will be a more formal handwritten exam.
  • Do not bring any electronic devices except a traditional calculator.
  • The exam is closed book. No materials are allowed, including the course website.
    • You may use blank scratch paper and a pen.
    • iPads and tablets are not allowed for this exam.
  • Please remember to write your name on the exam.

Topics

  1. Coverage: Lectures 2-34 for both finals, excluding:
    • LU decomposition in Lecture 10
    • Fourier material in Lecture 18
    • complex matrices in Lecture 30
    • the more technical details of SVD and PCA
  2. You should still understand the basic definitions and main conceptual ideas behind SVD and PCA.
  3. Final 1 will focus more on quick understanding checks, short computations, and True / False questions.
  4. Final 2 will emphasize deeper understanding and the relationships between concepts.
  5. I am still finalizing the exams, so I will share more detailed information before the next review class.

Questions

Question 1

Let \[ \vec{u}=\langle 2,-1 \rangle, \qquad \vec{v}=\langle 1,3 \rangle, \qquad A= \begin{bmatrix} 1&0&2\\ 2&-1&1 \end{bmatrix}, \qquad \vec{x}=\langle 1,-2,3 \rangle. \]

Compute:

  1. \(2\vec{u}-\vec{v}\)
  2. \(A\vec{x}\)
  3. the size of \(A^TA\)

Solution 1

\[ 2\vec{u}-\vec{v} = 2\langle 2,-1 \rangle-\langle 1,3 \rangle = \langle 4,-2 \rangle-\langle 1,3 \rangle = \langle 3,-5 \rangle. \]

\[ A\vec{x} = \begin{bmatrix} 1&0&2\\ 2&-1&1 \end{bmatrix} \begin{bmatrix} 1\\-2\\3 \end{bmatrix} = \begin{bmatrix} 1+0+6\\ 2+2+3 \end{bmatrix} = \langle 7,7 \rangle. \]

Also, \[ A^TA \text{ has size } 3\times 3. \]

Question 2

Suppose \[ T(\langle 1,0 \rangle)=\langle 1,2 \rangle, \qquad T(\langle 0,1 \rangle)=\langle -2,1 \rangle. \]

  1. Find \(T(\langle 3,-1 \rangle)\).
  2. Explain why knowing \(T(e_1)\) and \(T(e_2)\) is enough.

Solution 2

  • Since \(\langle 3,-1 \rangle=3e_1-e_2\), linearity gives \[ T(\langle 3,-1 \rangle)=3T(e_1)-T(e_2). \]
  • Therefore \[ T(\langle 3,-1 \rangle) = 3\langle 1,2 \rangle-\langle -2,1 \rangle = \langle 3,6 \rangle-\langle -2,1 \rangle = \langle 5,5 \rangle. \]
  • Every vector in \(\mathbb{R}^2\) is a linear combination of \(e_1\) and \(e_2\).
  • So a linear transformation is determined by its values on a basis.

Question 3

The RREF of an augmented matrix is \[ \left[ \begin{array}{ccc|c} 1&0&4&3\\ 0&1&-2&-1\\ 0&0&0&0 \end{array} \right]. \]

  1. Does the system have one solution, no solution, or infinitely many solutions?
  2. Write the solution in parametric form.

Solution 3

  • Column 3 has no pivot, so \(x_3\) is free.
  • Let \(x_3=t\).
  • Then \[ x_1+4x_3=3, \qquad x_2-2x_3=-1. \]
  • So \[ x_1=3-4t, \qquad x_2=-1+2t. \]
  • Hence the system has infinitely many solutions: \[ (x_1,x_2,x_3)=(3-4t,\ -1+2t,\ t). \]

Question 4

Let \[ A= \begin{bmatrix} 2&1\\ 5&3 \end{bmatrix}. \]

  1. Find \(A^{-1}\).
  2. State one fact about elementary matrices that helps with inverses.

Solution 4

For a \(2\times 2\) matrix, \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix}. \]

Here \[ \det(A)=2\cdot 3-1\cdot 5=1, \] so \[ A^{-1} = \begin{bmatrix} 3&-1\\ -5&2 \end{bmatrix}. \]

Also, every elementary matrix is invertible, and its inverse is again elementary.

Question 5

A two-state Markov chain has states \(A\) and \(B\).

  • 70% of state \(A\) stays in \(A\)
  • 40% of state \(B\) moves to \(A\)

Write the transition matrix \(P\).

Solution 5

\[ P= \begin{bmatrix} 0.7&0.4\\ 0.3&0.6 \end{bmatrix}. \]

Question 6

In \(P_2\), determine whether each set is a subspace:

  1. \(S=\{p(x)\mid p(0)+p(1)=0\}\)
  2. \(T=\{p(x)\mid p(0)+p(1)=1\}\)

Solution 6

  • \(S\) is a subspace:
    • the zero polynomial is in \(S\)
    • \(S\) is closed under addition
    • \(S\) is closed under scalar multiplication
  • \(T\) is not a subspace because the zero polynomial is not in \(T\).

Question 7

Suppose a \(4\times 5\) matrix has RREF \[ \begin{bmatrix} 1&0&3&0&2\\ 0&1&-1&0&4\\ 0&0&0&1&5\\ 0&0&0&0&0 \end{bmatrix}. \]

Find:

  1. the rank
  2. the nullity
  3. one basis vector for the null space

Solution 7

  • Pivot columns are \(1\), \(2\), and \(4\), so \[ \mathop{\mathrm{rank}}(A)=3. \]
  • There are \(5\) columns, so \[ \mathop{\mathrm{null}}(A)=5-3=2. \]
  • Set \(x_3=1\) and \(x_5=0\).
  • Then \[ x_1=-3,\qquad x_2=1,\qquad x_4=0. \]
  • One basis vector for the null space is \[ \langle -3,1,1,0,0 \rangle. \]

Question 8

Let \[ \vec{w}_1=\langle 1,0,2 \rangle, \qquad \vec{w}_2=\langle 0,1,-1 \rangle, \qquad \vec{w}_3=\langle 1,1,1 \rangle. \]

Are these vectors linearly independent?

Solution 8

Notice that \[ \vec{w}_1+\vec{w}_2 = \langle 1,1,1 \rangle = \vec{w}_3. \]

So one vector is a linear combination of the others, and the set is linearly dependent.

Question 9

Let \[ \vec{u}=\langle 1,2,2 \rangle, \qquad \vec{v}=\langle 2,0,1 \rangle, \qquad \vec{b}=\langle 4,1,1 \rangle. \]

  1. Compute \(\left\lVert \vec{u} \right\rVert\).
  2. Compute \(\vec{u}\cdot \vec{v}\) and decide whether the angle is acute, right, or obtuse.
  3. Find the projection of \(\vec{b}\) onto \(\mathop{\mathrm{span}}\{\langle 1,1,0 \rangle\}\).

Solution 9

First, \[ \left\lVert \vec{u} \right\rVert = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3. \]

Next, \[ \vec{u}\cdot \vec{v} = 1\cdot 2+2\cdot 0+2\cdot 1=4. \]

Since the dot product is positive, the angle is acute.

Let \(\vec{a}=\langle 1,1,0 \rangle\). Then \[ \operatorname{proj}_{\vec{a}}\vec{b} = \frac{\vec{b}\cdot \vec{a}}{\vec{a}\cdot \vec{a}}\vec{a} = \frac{4+1}{1+1}\vec{a} = \frac52 \vec{a} = \langle \frac 52,\frac 52,0 \rangle. \]

Question 10

Let \[ A= \begin{bmatrix} 1&0\\ 1&1\\ 1&2 \end{bmatrix}, \qquad \vec{b}= \begin{bmatrix} 1\\2\\2 \end{bmatrix}. \]

Write the normal equations for the least-squares problem \[ A\hat{\vec{x}}\approx \vec{b}. \]

Solution 10

The normal equations are \[ A^TA\hat{\vec{x}}=A^T\vec{b}. \]

Here \[ A^TA= \begin{bmatrix} 3&3\\ 3&5 \end{bmatrix}, \qquad A^T\vec{b}= \begin{bmatrix} 5\\ 6 \end{bmatrix}. \]

So the normal equations are \[ \begin{bmatrix} 3&3\\ 3&5 \end{bmatrix} \hat{\vec{x}} = \begin{bmatrix} 5\\ 6 \end{bmatrix}. \]

Question 11

Answer these quick checks.

  1. If \[ \det(A)=4,\qquad \det(A_1)=12,\qquad \det(A_2)=-8,\qquad \det(A_3)=0, \] find the solution to \(A\vec{x}=\vec{b}\) using Cramer’s rule.
  2. Compute \[ \langle 1,0,2 \rangle\times \langle 0,1,1 \rangle. \]
  3. If \(\det(B)=-3\) for a \(2\times 2\) matrix \(B\), what happens to area and orientation under \(x\mapsto Bx\)?

Solution 11

  • By Cramer’s rule, \[ x_1=\frac{12}{4}=3,\qquad x_2=\frac{-8}{4}=-2,\qquad x_3=\frac{0}{4}=0. \]
  • So \[ \vec{x}=\langle 3,-2,0 \rangle. \]
  • Also, \[ \langle 1,0,2 \rangle\times \langle 0,1,1 \rangle = \langle -2,-1,1 \rangle. \]
  • If \(\det(B)=-3\), then area scales by \(3\) and orientation reverses.

Question 12

Let \[ A= \begin{bmatrix} 5&2\\ 0&1 \end{bmatrix}. \]

  1. Find the eigenvalues of \(A\).
  2. Find one eigenvector corresponding to \(\lambda=1\).
  3. True or false: Eigenvectors corresponding to distinct eigenvalues are linearly independent.

Solution 12

  • Because \(A\) is upper triangular, the eigenvalues are the diagonal entries: \[ \lambda=5,\qquad \lambda=1. \]
  • For \(\lambda=1\), solve \[ (A-I)\vec{u}=\vec{0}. \]
  • Since \[ A-I= \begin{bmatrix} 4&2\\ 0&0 \end{bmatrix}, \] we get \(4x+2y=0\), so \(y=-2x\).
  • One eigenvector is \[ \langle 1,-2 \rangle. \]
  • The statement is true.

Question 13

Answer these quick later-course questions.

  1. Solve \[ \vec{x}'(t)=D\vec{x}(t), \qquad D= \begin{bmatrix} 2&0\\ 0&-1 \end{bmatrix}, \qquad \vec{x}(0)=\langle 3,1 \rangle. \]
  2. True or false: Every real matrix has an SVD.
  3. True or false: The number of nonzero singular values equals the rank.
  4. In PCA, what does the first principal direction represent?

Solution 13

  • Because \(D\) is diagonal, \[ \vec{x}(t)=\langle 3e^{2t},e^{-t} \rangle. \]
  • Every real matrix has an SVD: True.
  • The number of nonzero singular values equals the rank: True.
  • In PCA, the first principal direction is the direction of largest variance in the data.

Question 14

Decide whether each statement is true or false.

  1. Every elementary matrix is invertible.
  2. If \(Ax=\vec{0}\) has only the trivial solution, then the columns of \(A\) are linearly independent.
  3. If \(\det(A)=0\), then \(A\) is invertible.
  4. Every square matrix is diagonalizable.
  5. The number of nonzero singular values equals the rank.

Solution 14

    1. True
    1. True
    1. False
    1. False
    1. True