# MATH 125 Final: Sample Final Exam - Solutions

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University of Saskatchewan

Mathematics

MATH 125

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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
2 3
0 ▯1 0 2
6 7
Q = 6 ▯1 2 0 1 7
4 2 ▯3 2 4 5
0 0 3 2
Solution :
▯ ▯ ▯ ▯
▯ 0 ▯1 0 2 ▯ ▯ 0 ▯1 0 2 ▯
▯ ▯1 2 0 1 ▯ ▯ ▯1 2 0 1 ▯
det(Q) = ▯ ▯= ▯ ▯ (3 + 2r2)
▯ 2 ▯3 2 4 ▯ ▯ 0 1 2 6 ▯
▯ 0 0 3 2 ▯ ▯ 0 0 3 2 ▯
▯ ▯
▯ ▯
▯▯1 0 2 ▯
= ▯(▯1) ▯ 1 2 6 ▯ (expand along 1st column)
▯ 0 3 2 ▯
▯ ▯
▯0 2 8 ▯
▯ ▯
= ▯1 2 6 ▯ (r1+ r2)
▯0 3 2 ▯
▯ ▯
▯ 2 8 ▯
= ▯1 ▯ 3 2 ▯ (expand along 1st column)
= ▯1(▯20) = 20:
2. Consider the matrices
2 3 2 3
0 4 8 4 ▯24 0 1 0 ▯3 0
60 ▯3 ▯6 ▯3 18 7 60 0 1 2 07
6 7 6 7
A = 60 2 4 2 ▯12 7 and B = 60 0 0 0 17
40 ▯2 ▯3 0 7 5 40 0 0 0 05
0 3 4 ▯1 ▯5 0 0 0 0 0
Given that B is the reduced row echelon form of A: (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.
5
(e) (2 points) Do the columns of A form a basis for R ? Justify your answer.
Solution: (a) The non-zero rows of B form a basis of the row space of A:
▯h i h i h i▯
0 1 0 ▯3 0 ; 0 0 1 2 0 ; 0 0 0 0 1 :
(b) The columns of A corresponding to those columns of B which contain the leading 1’s form
a basis of the column space of A:
8 9
> 2 3 2 3 2 3>
> 4 8 ▯24 >
< 6▯3 7 6 ▯67 6 187=
6 27;6 47 ; ▯12 7
> 6 7 6 7 6 7>
> 4▯2 5 4 ▯35 4 7 5>
: 3 4 ▯5 ;
(c) The reduced row echelon form of the augmented matrix of tx = 0mogeneous system A~
is equal to [Bj0], i.e.,
2 3
0 1 0 ▯3 0 0
6 7
6 0 0 1 2 0 07
[Aj0] ▯! ▯▯▯6▯! 0 0 0 1 07
4 0 0 0 0 0 05
0 0 0 0 0 0
Thus, there are three leading variables i2 3his s5stem (x , x and x ), and the remaining
variables (x ;x ) are free. Assigning the parametric values s and t to the free variables x and
1 4 ~ 1
x4, respectively, we get that the genx = 0 isution of A~
2 3 2 3
1 0
60 7 6 37
6 7 6 7
x = 60 7+ t6 ▯27 (s;t 2 R):
40 5 4 15
0 0
Hence
8 2 3 2 3 9
> 1 0 >
>6 07 6 37>>
<66 7 6 7=
>66 07 6 ▯27>
> 4 05 4 15 >
: 0 0 ; is a basis of null(A).
(d) rank(A) = 3 (B has three non-zero rows), and nullity(A) = dim(null(A)) = 2.
5
(e) No, the columns of A do not form a basis of R . There are many ways to argue here. The
easiest is to note that the ▯rst column of A is the zero vector, and a collection of vectors which
5
includes the zero vector cannot be linearly independent, and therefore cannot be a basis of R
(a basis is required to be a linearly independent set by de▯nition).
Alternatively, we can appeal to a general result from lectures: The columns of an n ▯ n
matrix A form a basis of R if and only if A is invertible. But A is 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 R
3. Let 2 3
5 ▯6 8
6 7
A = 4 0 3 05
0 0 3
(a) (3 points) Find the characteristic polynomial of A. Show your work.
(b) (8 points) The eigenvalues of A are 3 and 5 . Find a basis for the eigenspace E 3f A
and a basis for the eigenspace 5 of A.
(c) (2 points) Diagonalize A, that is, ▯nd an invertible matrix P and a diagonal matrix D
▯1 ▯1
such that P AP = D: (Do not compute P !)
Solution : (a)
▯ ▯
▯ 5 ▯ ▯ ▯6 8 ▯
CA(▯) = ▯ 0 3 ▯ ▯ 0 ▯
▯ ▯
▯ 0 0 3 ▯ ▯ ▯
▯ ▯
▯ 3 ▯ ▯ 0 ▯
= (5 ▯ ▯)▯ 0 3 ▯ ▯ ▯ (expand along 1st column)
2
= (5 ▯ ▯)(3 ▯ ▯)
(this can be rewritten as A (▯) = ▯▯ + 11▯ ▯ 39▯ + 45).
(b) The eigenspace E is the null space of the matrix A▯3I . We solve the homogeneous linear
3 3
system (A ▯ 3I 3x = 0:
2 3 2 3
2 ▯6 8 0 r1▯ 2 1 ▯3 4 0
[A ▯ 3I j0] =6 0 0 0 0 7 ▯! 6 0 0 0 0 7
3 4 5 4 5
0 0 0 0 0 0 0 0
From the reduced row echelon form, we see that the general solution of the system is 2 3 2 3
3 ▯4
6 7 6 7
x = 4 1 5+ t4 05 (s;t 2 R);
0 1
and so a basis3of E = nul3(A ▯ 3I ) is given by
82 3 2 39
< 3 ▯4 =
B = 6 17 ; 07 :
3 >4 5 4 5>
: 0 1 ;
The eigenspac5 E is the null space of the 3atrix A ▯ 5I . We solve the homogeneous linear
system (A ▯3x = 0:
2 3 2 3
0 ▯6 8 0 0 1 0 0
[A ▯ 5I j0] = ▯2 0 0 7▯! ▯▯▯ ▯!6 0 0 1 07
3 4 5 4 5
0 0 ▯2 0 0 0 0 0
From the reduced row echelon form, we see that the general solution of the system is
2 3
1
6 7
x = 4 0 5 (t 2 R);
0
and so a basis5of E = nul3(A ▯ 5I ) is given by
8 2 39
< 1 =
6 7
B5= > 40 5> :
: 0 ;
(c) Note that A is diagonalizable since the algebraic and geometric multiplicities agree for both
eigenvalues. By part (b), we have
P ▯AP = D;
where
2 3 2 3
6 3 ▯4 1 7 6 3 0 0 7
P = 4 1 0 0 5 and D = 4 0 3 0 5
0 1 0 0 0 5
(so the i-th column of P is the i-th eigenvector of A found when computing bases of the
eigenspaces in part (b), and the i-th diagonal entry of D is its corresponding eigenvalue).
Remark: This is not the only correct answer. Di▯erent choices of bases of the eigenspaces and
di▯erent orderings of the eigenvectors can lead to di▯erent matrices P and D. You can verify that your answer is correct by checking that AP = PD.
4. (a) (9 points) Find the inverse of the matrix
2 3
6 1 ▯2 ▯4 7
P = 4 0 1 2 5
2 ▯1 ▯1
" # " #
▯5 ▯3 0 ▯3
(b) (5 points) Let C = 3 2 and B = 1 6 . Find the invertible matrix X such
that ▯ ▯▯1
X ▯C ▯1 + 4X = B T
(c) (2 points) Let x = b be a system of linear equations such that A is 4 ▯ 3. Suppose
~ ~
the augmented matrix [ A j is invertible. Dx = b have a solution? Justify your
answer.
Solution (a) We use the standard algorithm:
2 3 2 3
1 ▯2 ▯4 1 0 0 1 0 0 1 2 0
6 7 6 7
4 0 1 2 0 1 0 5 ▯! ▯▯▯ ▯! 4 0 1 0 4 7 ▯2 5
2 ▯1 ▯1 0 0 1 0 0 1 ▯2 ▯3 1
Since the reduced row echelon form o3 P is I (the left-hand side of the partitioned matrix), P
is invertible and 2 3
1 2 0
▯1 6 7
P = 4 4 7 ▯2 5
▯2 ▯3 1
(the right-hand side of the partitioned matrix).
▯ ▯
▯1 ▯1 ▯1 ▯1 ▯1 ▯1 ▯1
(b) SinceX C = (C ) (X ) = CX, we have that
▯ ▯▯1
X▯1C ▯1 + 4X = B T ) CX + 4X = B T
) (C + 4I 2X = B T (i)
Now,
" # " # " #
▯5 ▯3 4 0 ▯1 ▯3
C + 42 = 3 2 + 0 4 = 3 6 ;
In particular, det(C

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