Math 369 Exam #1 Practice Problem Solutions
1. Find the set of solutions of the following system of linear equations. Show enough work to make your
steps clear.
x + 2y + 3z + 4w = 1
−2x − 3y − 4z − 6w = 1
3x + 5y + 7z + 10w = 0
Answer: We solve by forming the augmented matrix

1
2
3
4
−2 −3 −4 −6
3
5
7
10

1
1
0
and row-reducing. First, replace row 2 with row 2 plus twice row 1 and replace row 3 with row 3 minus
3 times row 1 to get


1 2
3
4
1
0 1
2
2
3
0 −1 −2 −2 −3
Then replace row 3 with row 3 plus row 2 and replace row 1 with row 1 minus twice row 2:


1 0 −1 0 −5
0 1 2 2 3 
0 0 0 0 0
Therefore, the solutions of the system are
x=z−5
y = 3 − 2z − 2w,
where z and w are free variables.


a 0 0
2. (a) Suppose a, b, c are nonzero numbers. Find the inverse of the matrix 0 b 0.
0 0 c


1/a
0
0
Answer: I claim that  0 1/b 0  is the inverse. To see this, I just need to multiply the two
0
0 1/c
matrices both ways and see that I get the identity matrix:


 

1/a
a 0 0
0
0
1 0 0
0 b 0  0 1/b 0  = 0 1 0
0 0 c
0
0 1/c
0 0 1


 

1/a
0
0
a 0 0
1 0 0
 0 1/b 0  0 b 0 = 0 1 0
0
0 1/c
0 0 c
0 0 1

1
(b) Find the inverse of the matrix 0
0

d e
1 f , where d, e, f are any real numbers.
0 1
1
Answer: To find the inverse, I form the

1
0
0
augmented matrix

d e 1 0 0
1 f 0 1 0
0 1 0 0 1
and row-reduce. First, subtract d times row 2 from row 1

1 0 e − df 1 −d
0 1
0 1
f
0 0
1
0 0
Then subtract (e − df ) times row 3 from

1 0
0 1
0 0
to get

0
0
1
row 1 and subtract f times row 3 from row 2 to get

0 1 −d df − e
0 0 1
−f 
1
1 0 0
Since the left half of the augmented matrix is the identity, the right half must be the inverse of
the given matrix, so we see that the inverse is


1 −d df − e
0 1
−f 
0 0
1
(c) True or false: Every upper triangular matrix with nonzero diagonal entries has an inverse. Explain
your answer.
Hint: Every such matrix can be written as the product

 


a x y
a 0 0
1 x/a y/a
0 b z  = 0 b 0 0 1 z/b 
0 0 c
0 0 c
0 0
1


a x y
Answer: True. To see this, let A = 0 b z  be an arbitrary
0 0 c
as described in the hint, A = BC where



1 x/a
a 0 0



and C = 0 1
B= 0 b 0
0 0
0 0 c
upper-triangular matrix. Then,

y/a
z/b  .
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We saw in parts (a) and (b) that B and C are invertible, so by HW 3 #4, we know that A−1 =
C −1 B −1 , so A is also invertible.
3. Let A ∈ Mat2,3 (R) be a 2 × 3 matrix.
*
*
*
(a) Let U = { x ∈ R3 : A x = 0 }. Show that U is a subspace of R3 .
*
* *
*
*
*
Proof. It suffices to show that (i) 0 ∈ U ; (ii) if x, y ∈ U , then x + y ∈ U ; and (iii) if x ∈ U and
*
λ ∈ R, then λ x ∈ U .
*
*
*
i. Since A 0 = 0 , we see immediately that 0 ∈ U .
2
* *
*
*
*
*
ii. Suppose x, y ∈ U . Then A x = 0 and A y = 0 . But then
*
*
*
*
*
*
*
*
*
A( x + y ) = A x + A y = 0 + 0 = 0 ,
*
*
so x + y ∈ U .
*
*
*
iii. Suppose x ∈ U and λ ∈ R. Then A x = 0 and so
*
*
A(λ x) = λA x = λ · 0 = 0 ,
*
so λ x ∈ U .
Having proved (i), (ii), and (iii), we conclude that U is indeed a subspace of R3 .
*
*
1
*
*
3
(b) Is W = { x ∈ R : A x = b } a subspace when b =
? Explain.
2
*
Answer: No. In particular, 0 ∈
/ W since
*
*
*
A 0 = 0 6= b ,
so W cannot be a subspace.
4. Let V be a vector space.
(a) Define what it means for a set {u1 , . . . , un } ⊂ V to be linearly dependent.
Answer: By definition, {u1 , . . . , un } is linearly dependent if there is a non-trivial way to write 0
as a linear combination of the ui , meaning that there exist scalars λ1 , . . . , λn not all zero so that
n
X
λi ui = 0.
i=1
(b) Suppose v ∈ V . Is the set {0, v} linearly dependent? Explain.
Answer: Yes. Let λ be any nonzero scalar. Then
λ·0+0·v =0
is a nontrivial linear combination of 0 and v that yields 0, so this set is linearly dependent.
5. Let V = P(R)[z] be the vector space of polynomials with real coefficients in the variable z. Is the set
{1 + z + z 2 , 1 − z, 1 − z 3 }
linearly dependent or linearly independent? Prove your claim.
Answer: This set is linearly independent. To see this, suppose a, b, c ∈ R are constants so that
a(1 + z + z 2 ) + b(1 − z) + c(1 − z 3 ) = 0.
Then, after distributing and combining terms, we see that
(a + b + c) + (a − b)z + az 2 − cz 3 = 0 + 0z + 0z 2 + 0z 3 .
Since the coefficients of z 2 and z 3 must be zero, we see that a = 0 and c = 0. But then the above
reduces to
b − bz = 0 + 0z + 0z 2 + 0z 3 ,
so b = 0 as well. Therefore, the only way to write 0 as a linear combination of 1 + z + z 2 , 1 − z, and
1 − z 3 is if all the coefficients are zero, which means the set is linearly independent.
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