Final Review II

Lecture 39

Minjae Park

Auburn University
MATH 2660 - Spring 2026

April 22, 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

Coverage is still Lectures 2-34, excluding:
- LU decomposition in Lecture 10
- Fourier material in Lecture 18
- complex matrices in Lecture 30
- the more technical details of SVD and PCA

What To Expect

Final 1

Final 1 is adapted heavily from Quizzes 1, 2, and 3.
Many questions will be very close in style to those quiz problems.
So your main study source should be the past quizzes, together with the review slides and review questions.
There will be 7 questions plus 1 TF question.

Final 2

Final 2 will have 5 longer problems, and each problem will have 4 connected parts.
The parts are meant to fit together rather than feel like four unrelated mini-problems.
So one problem may ask for both:
- a computation
- a short explanation of what that computation means
Clear reasoning and interpretation will matter, not just the final numeric answer.
Today’s review questions are not copies of the final questions. They are different practice problems built around the same themes and level of connection.

Final 2 (continued)

For Final 2, the main emphasis will be these five themes:
- linear transformations and determinants
- RREF and fundamental subspaces
- least squares and projections
- elementary matrices and Gauss-Jordan elimination
- diagonalization and linear ODEs
These themes are naturally connected to other concepts. For example, diagonalization involves eigenvalues and eigenvectors, and projections involve dot products and orthogonality.

Questions

Question 1

Suppose a linear transformation \(T:\mathbb{R}^2\to\mathbb{R}^2\) satisfies \[ T(e_1)=\langle 2,1 \rangle, \qquad T(e_2)=\langle -1,2 \rangle. \]

Answer the following.

Write the matrix of \(T\).
Compute \(T(\langle 1,3 \rangle)\).
Compute \(\det(T)\) and explain what it tells you about area, orientation, and invertibility.
Find the vector \(\vec{x}\) such that \[ T(\vec{x})=\langle 4,5 \rangle. \]

Solution 1

Since the columns of the matrix are the images of \(e_1\) and \(e_2\), we get \[ A= \begin{bmatrix} 2&-1\\ 1&2 \end{bmatrix}. \]
Therefore \[ T(\langle 1,3 \rangle) = \begin{bmatrix} 2&-1\\ 1&2 \end{bmatrix} \langle 1,3 \rangle = \langle -1,7 \rangle. \]
Next, \[ \det(T)=\det(A)=2\cdot 2-(-1)\cdot 1=5. \]
So areas are multiplied by \(5\), orientation is preserved, and the transformation is invertible.
To solve \[ A\vec{x}=\langle 4,5 \rangle, \] use \[ A^{-1} = \frac{1}{5} \begin{bmatrix} 2&1\\ -1&2 \end{bmatrix}. \]
Hence \[ \vec{x} = A^{-1}\langle 4,5 \rangle = \frac{1}{5} \begin{bmatrix} 2&1\\ -1&2 \end{bmatrix} \langle 4,5 \rangle = \langle \frac{13}{5},\frac 65 \rangle. \]

Question 2

Suppose a \(4\times 5\) matrix \(B\) has \[ \mathrm{RREF}(B)= \left[ \begin{array}{ccccc} 1&0&2&-1&0\\ 0&1&-3&4&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{array} \right]. \]

Let \(\vec{b}_1,\dots,\vec{b}_5\) denote the original columns of \(B\).

Answer the following.

Find a basis for \(\mathrm{Col}(B)\) and a basis for \(\mathrm{Row}(B)\).
Find the rank, nullity, and left nullity of \(B\).
Find a basis for the nullspace \(N(B)\).
Decide whether \(\{\vec{b}_1,\vec{b}_2,\vec{b}_5\}\) is a basis for \(\mathrm{Col}(B)\), and explain whether \(\vec{b}_4\) belongs to the span of the pivot columns.

Solution 2

The pivot columns are \(1\), \(2\), and \(5\).
Therefore a basis for the column space is \[ \{\vec{b}_1,\vec{b}_2,\vec{b}_5\}. \]
A basis for the row space can be read from the nonzero rows of the RREF: \[ \left\{ \langle 1,0,2,-1,0 \rangle, \langle 0,1,-3,4,0 \rangle, \langle 0,0,0,0,1 \rangle \right\}. \]
Therefore \[ \mathop{\mathrm{rank}}(B)=3. \]
Since there are \(5\) columns, \[ \mathop{\mathrm{null}}(B)=5-3=2. \]
Since there are \(4\) rows, \[ \text{left nullity}=4-3=1. \]
To find the nullspace, solve \[ \mathrm{RREF}(B)\vec{x}=\vec{0}. \]
Let \[ x_3=s, \qquad x_4=t. \]
Then \[ x_1=-2s+t, \qquad x_2=3s-4t, \qquad x_5=0. \]
So \[ \vec{x} = s\langle -2,3,1,0,0 \rangle + t\langle 1,-4,0,1,0 \rangle. \]
A basis for \(N(B)\) is \[ \left\{ \langle -2,3,1,0,0 \rangle, \langle 1,-4,0,1,0 \rangle \right\}. \]
The set \[ \{\vec{b}_1,\vec{b}_2,\vec{b}_5\} \] is a basis for \(\mathrm{Col}(B)\) because these are exactly the original pivot columns.
Also, \(\vec{b}_4\) lies in the span of the pivot columns because column \(4\) is not a pivot column.
In fact, the RREF gives the relation \[ \vec{b}_4=-\vec{b}_1+4\vec{b}_2. \]

Question 3

Let \[ \vec{u}=\langle 1,1,0 \rangle, \qquad \vec{v}=\langle 1,-1,0 \rangle, \qquad \vec{b}=\langle 3,1,2 \rangle, \] and let \[ A=[\vec{u}\ \vec{v}]. \]

Answer the following.

Show that \(\vec{u}\) and \(\vec{v}\) are orthogonal, and compute their norms.
Write the normal equations for the least-squares problem \[ A\hat{\vec{x}}\approx \vec{b}. \]
Find the least-squares solution \(\hat{\vec{x}}\) and the projection \(\hat{\vec{b}}\) of \(\vec{b}\) onto \(\mathrm{Col}(A)\).
Find the residual \(\vec{r}=\vec{b}-\hat{\vec{b}}\) and explain why it is orthogonal to the column space.

Solution 3

First, \[ \vec{u}\cdot \vec{v} = 1\cdot 1+1\cdot (-1)+0\cdot 0 = 0. \]
So the vectors are orthogonal.
Also, \[ \left\lVert \vec{u} \right\rVert=\left\lVert \vec{v} \right\rVert=\sqrt{2}. \]
The normal equations are \[ A^TA\hat{\vec{x}}=A^T\vec{b}. \]
Since the columns are orthogonal, \[ A^TA= \begin{bmatrix} 2&0\\ 0&2 \end{bmatrix}, \qquad A^T\vec{b}= \langle 4,2 \rangle. \]
So \[ \hat{\vec{x}} = \langle 2,1 \rangle. \]
Therefore \[ \hat{\vec{b}} = A\hat{\vec{x}} = 2\vec{u}+\vec{v} = \langle 3,1,0 \rangle. \]
The residual is \[ \vec{r} = \vec{b}-\hat{\vec{b}} = \langle 0,0,2 \rangle. \]
This residual is orthogonal to the column space because \[ \vec{r}\cdot \vec{u}=0, \qquad \vec{r}\cdot \vec{v}=0. \]
So \(\hat{\vec{b}}\) is the orthogonal projection of \(\vec{b}\) onto \(\mathrm{Col}(A)\).

Question 4

Let \[ M= \begin{bmatrix} 1&2\\ 3&7 \end{bmatrix}. \]

Answer the following.

Let \(E_1\) correspond to the row operation \(R_2\to R_2-3R_1\), and let \(E_2\) correspond to \(R_1\to R_1-2R_2\). Find \(E_1\) and \(E_2\).
Verify that \[ E_2E_1M=I. \]
Use this to find \(M^{-1}\).
Use elementary matrices to explain why \(\det(M)=1\).

Solution 4

Apply the first row operation to \(I_2\): \[ E_1= \begin{bmatrix} 1&0\\ -3&1 \end{bmatrix}. \]
Apply the second row operation to \(I_2\): \[ E_2= \begin{bmatrix} 1&-2\\ 0&1 \end{bmatrix}. \]
Then \[ E_1M = \begin{bmatrix} 1&0\\ -3&1 \end{bmatrix} \begin{bmatrix} 1&2\\ 3&7 \end{bmatrix} = \begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}, \]
and so \[ E_2E_1M = \begin{bmatrix} 1&-2\\ 0&1 \end{bmatrix} \begin{bmatrix} 1&2\\ 0&1 \end{bmatrix} = I. \]
Therefore \[ M^{-1}=E_2E_1 = \begin{bmatrix} 1&-2\\ 0&1 \end{bmatrix} \begin{bmatrix} 1&0\\ -3&1 \end{bmatrix} = \begin{bmatrix} 7&-2\\ -3&1 \end{bmatrix}. \]
Since both row operations are row replacements, \[ \det(E_1)=1, \qquad \det(E_2)=1. \]
Using \[ E_2E_1M=I, \] take determinants: \[ \det(E_2)\det(E_1)\det(M)=1. \]
So \[ \det(M)=1. \]
More generally, Gauss-Jordan elimination can be described by multiplying on the left by elementary matrices until the original matrix becomes the identity.

Question 5

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

Answer the following.

Find the eigenvalues and one eigenvector for each eigenvalue.
Build a diagonalization \[ A=PDP^{-1}. \]
Solve the system \[ \vec{x}'(t)=A\vec{x}(t), \qquad \vec{x}(0)=\langle 2,0 \rangle. \]
Describe the long-term behavior of \(\vec{x}(t)\) and explain which eigendirection dominates.

Solution 5

Compute the characteristic polynomial: \[ \det(A-\lambda I) = \begin{vmatrix} 3-\lambda&1\\ 1&3-\lambda \end{vmatrix} = (3-\lambda)^2-1. \]
So \[ (3-\lambda)^2-1=0 \quad \Rightarrow \quad \lambda=4,\ 2. \]
For \(\lambda=4\), an eigenvector is \[ \vec{u}_1=\langle 1,1 \rangle. \]
For \(\lambda=2\), an eigenvector is \[ \vec{u}_2=\langle 1,-1 \rangle. \]
Therefore \[ P= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \qquad D= \begin{bmatrix} 4&0\\ 0&2 \end{bmatrix}. \]
Now write the initial vector as \[ \langle 2,0 \rangle = \langle 1,1 \rangle+\langle 1,-1 \rangle = \vec{u}_1+\vec{u}_2. \]
So the solution is \[ \vec{x}(t) = e^{4t}\vec{u}_1+e^{2t}\vec{u}_2 = e^{4t}\langle 1,1 \rangle+e^{2t}\langle 1,-1 \rangle. \]
Equivalently, \[ \vec{x}(t)= \langle e^{4t}+e^{2t},\ e^{4t}-e^{2t} \rangle. \]
As \(t\to\infty\), the \(e^{4t}\vec{u}_1\) term dominates.
So the solution eventually points mostly in the direction of \[ \langle 1,1 \rangle. \]
In other words, the eigendirection for the larger eigenvalue dominates the long-term behavior.