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MATH 125 Study Guide - Final Guide: Row Echelon Form, Invertible Matrix, Diagonal Matrix


Department
Mathematics
Course Code
MATH 125
Professor
Scott
Study Guide
Final

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MATH 125 (R1) Winter 2017
Sample Final Exam – Solutions
Notes: Some steps are skipped in the solutions below (e.g., some row reduction of matrices is
omitted). In your exam, you should show all your work! Remember also that there are often a
number of ways to solve a given problem; here I have generally just given one method.
1. (9 points) Compute the determinant of Q, where
Q=
01 0 2
1 2 0 1
23 2 4
0 0 3 2
Solution:
det(Q) =
01 0 2
1 2 0 1
23 2 4
0 0 3 2
=
01 0 2
1 2 0 1
0 1 2 6
0 0 3 2
(r3+ 2r2)
=(1)
1 0 2
1 2 6
0 3 2
(expand along 1st column)
=
0 2 8
1 2 6
0 3 2
(r1+r2)
=1
2 8
3 2
(expand along 1st column)
=1(20) = 20.
2. Consider the matrices
A=
0 4 8 4 24
0363 18
0 2 4 2 12
023 0 7
0 3 4 15
and B=
0 1 0 3 0
0 0 1 2 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
Given that Bis the reduced row echelon form of A:

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(a) (3 points) Find a basis for row(A), the row space of A.
(b) (3 points) Find a basis for col(A), the column space of A.
(c) (4 points) Find a basis for null(A), the null space of A.
(d) (2 points) Find the rank and nullity of A.
(e) (2 points) Do the columns of Aform a basis for R5? Justify your answer.
Solution: (a) The non-zero rows of Bform a basis of the row space of A:
h0 1 0 3 0 i,h00120i,h00001i.
(b) The columns of Acorresponding to those columns of Bwhich contain the leading 1’s form
a basis of the column space of A:
4
3
2
2
3
,
8
6
4
3
4
,
24
18
12
7
5
(c) The reduced row echelon form of the augmented matrix of the homogeneous system A~x =~
0
is equal to [B|~
0], i.e.,
[A|~
0] → ··· −
0 1 0 3 0 0
0 0 1 2 0 0
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
Thus, there are three leading variables in this system (x2,x3and x5), and the remaining
variables (x1, x4) are free. Assigning the parametric values sand tto the free variables x1and
x4, respectively, we get that the general solution of A~x =~
0 is
~x =s
1
0
0
0
0
+t
0
3
2
1
0
(s, t R).
Hence
1
0
0
0
0
,
0
3
2
1
0

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is a basis of null(A).
(d) rank(A) = 3 (Bhas three non-zero rows), and nullity(A) = dim(null(A)) = 2.
(e) No, the columns of Ado not form a basis of R5. There are many ways to argue here. The
easiest is to note that the first column of Ais the zero vector, and a collection of vectors which
includes the zero vector cannot be linearly independent, and therefore cannot be a basis of R5
(a basis is required to be a linearly independent set by definition).
Alternatively, we can appeal to a general result from lectures: The columns of an n×n
matrix Aform a basis of Rnif and only if Ais invertible. But Ais not invertible here, since
(for example), nullity(A)6= 0 by part (d). Hence,as before, we conclude that the columns of A
do not form a basis of R5
3. Let
A=
56 8
030
003
(a) (3 points) Find the characteristic polynomial of A. Show your work.
(b) (8 points) The eigenvalues of Aare 3 and 5 . Find a basis for the eigenspace E3of A
and a basis for the eigenspace E5of A.
(c) (2 points) Diagonalize A, that is, find an invertible matrix Pand a diagonal matrix D
such that P1AP =D. (Do not compute P1!)
Solution: (a)
CA(λ) =
5λ6 8
0 3 λ0
0 0 3 λ
= (5 λ)
3λ0
0 3 λ
(expand along 1st column)
= (5 λ)(3 λ)2
(this can be rewritten as CA(λ) = λ3+ 11λ239λ+ 45).
(b) The eigenspace E3is the null space of the matrix A3I3. We solve the homogeneous linear
system (A3I3)~x =~
0:
[A3I3|~
0] =
26 8 0
0 0 0 0
0 0 0 0
r1×1
2
13 4 0
0 0 0 0
0 0 0 0
From the reduced row echelon form, we see that the general solution of the system is
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