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MATH 1553
PRACTICE MIDTERM 3 (VERSION A)
Name (^) GT Email @gatech.edu
Please read all instructions carefully before beginning.
- Please leave your GT ID card on your desk until your TA matches your exam.
- Each problem is worth 10 points. The maximum score on this exam is 50 points.
- You have 50 minutes to complete this exam.
- There are no aids of any kind (notes, text, etc.) allowed.
- Please show your work.
- You may cite any theorem proved in class or in the sections we covered in the text.
- Good luck!
This is a practice exam. It is similar in format, length, and difficulty to the real exam. It is not meant as a comprehensive list of study prob- lems. I recommend completing the practice exam in 50 minutes, with- out notes or distractions.
The exam is not designed to test material from the previous midterm on its own. However, knowledge of the material prior to chapter 3 is necessary for everything we do for the rest of the semester, so it is fair game for the exam as it applies to chapters 3 and 5.
We’ll be using an online system to help grade the exams this time, and the QR codes allow the scanner to mach each page to you. Don’t mess them up!
Problem 1. [2 points each]
In this problem, if the statement is always true, circle T ; if it is always false, circle F ; if it is sometimes true and sometimes false, circle M.
a) T F M If A is a 3 × 3 matrix with characteristic polynomial − λ^3 +
λ^2 + λ , then A is invertible.
b) T F M A 3 × 3 matrix with (only) two distinct eigenvalues is di-
agonalizable.
c) T F M If A is diagonalizable and B is similar to A , then B is diag-
onalizable.
d) T F M A diagonalizable n × n matrix admits n linearly indepen-
dent eigenvectors.
e) T F M If det( A ) = 0, then 0 is an eigenvalue of A.
Solution.
a) False: λ = 0 is a root of the characteristic polynomial, so 0 is an eigenvalue, and A is not invertible. b) Maybe: it is diagonalizable if and only if the eigenspace for the eigenvalue with multiplicity 2 has dimension 2. c) True: if A = P DP −^1 with D diagonal, and B = CAC −^1 , then B = C ( P DP −^1 ) C −^1 = ( C P ) D ( C P )−^1 , so B is also similar to a diagonal matrix. d) True: by the Diagonalization Theorem, an n × n matrix is diagonalizable if and only if it admits n linearly independent eigenvectors. e) True: if det( A ) = 0 then A is not invertible, so Av = 0 v has a nontrivial solution.
Problem 3.
Consider the matrix
A =
a) [4 points] Find the eigenvalues of A , and compute their algebraic multiplici- ties. b) [4 points] For each eigenvalue of A , find a basis for the corresponding eigenspace. c) [2 points] Is A diagonalizable? If so, find an invertible matrix P and a diagonal matrix D such that A = P DP −^1. If not, why not?
Solution.
a) We compute the characteristic polynomial by expanding along the second row:
f ( λ ) = det
4 − λ 2 − 4 0 2 − λ 0 2 2 − 2 − λ
= ( 2 − λ ) det
Å
4 − λ − 4 2 − 2 − λ
ã
= ( 2 − λ )( λ^2 − 2 λ ) = − λ ( λ − 2 )^2 The roots are 0 (with multiplicity 1) and 2 (with multiplicity 2). b) First we compute the 0-eigenspace by solving ( A − 0 I ) x = 0:
A =
rref 1 0 −^1 0 1 0 0 0 0
The parametric vector form of the general solution is
x y z
= z
, so a basis for
the 0-eigenspace is
Next we compute the 2-eigenspace by solving ( A − 2 I ) x = 0:
A − 2 I =
rref 1 1 −^2 0 0 0 0 0 0
The parametric vector form for the general solution is
x y z
= y
, so
a basis for the 2-eigenspace is
c) We have produced three linearly independent eigenvectors, so the matrix is diago- nalizable:
A =
Problem 5. [10 points]
Compute the determinant of the matrix
Solution.
The determinant of the cube is the cube of the determinant, so we start by computing
det
This is a big, complicated matrix, so it’s easiest to use row reduction.
det
=^ det
(row replacements)
= det
(row replacements)
= − det
(row swap)
= − 1 · 1 · 1 · 10 (triangular matrix) = −10.
Thus
det
[Scratch work]
