Linear Algebra
Final Exam Review
The Final Exam will be on Tuesday, May 9, 8:00am-10:30am.
25% of your overall grade is based on the Final Exam. If your Final Exam grade is higher than one of your
checkpoint grades, then the Final Exam grade will also replace one low checkpoint exam grade and therefore count
for an additional 12% in your overall grade.
Linear Algebra Learning Objectives: Students should be able to...
Write a coefficient and/or augmented matrix and put it in REF or RREF using elementary row operations.
Identify the pivot positions and pivot columns in a matrix; identify the basic variables and free variables in a
linear system and parameterize the solution (when a solution exists) accordingly.
Write the solutions of a non-homogeneous linear system Ax = b as x = vh + p where vh is the solution set
to the homogeneous equation Ax = 0 and p is any particular solution to the non-homogeneous equation.
Determine whether a given vector is in the span of a set of vectors (in Rn , Pn , or Mm×n ), and if so, find the
linear combinations that produce the given vector.
Determine whether a set of vectors (in Rn , Pn , or Mm×n ) is linearly independent and, if not, find the linear
dependence relation.
Perform addition, scalar multiplication, matrix multiplication, and transposition on matrices.
Determine whether the inverse of a matrix exists, and if so, find it.
Compute the determinant of a square matrix (up to 3×3) using cofactor expansion or any relevant determinant
properties.
Determine whether a set of vectors constitutes a subspace of Rn , Pn , or Mm×n .
Determine whether a transformation between two vector spaces is linear, and whether it is one-to-one and/or
onto. Find the kernel and range of a linear transformation.
Determine and prove whether a set acts as a basis for a vector space, and find the dimension of a vector space
or subspace.
Find bases for the null space and column space of a matrix, and find the rank of a matrix.
Find the characteristic polynomial, eigenvalues, and corresponding eigenvectors of a matrix, and the algebraic
multiplicity and geometric multiplicities of the eigenvalues.
Apply the Invertible Matrix Theorem.
State and apply the definitions: REF, RREF, pivot position/column, linear combination, span, homogeneous
linear system, trivial solution, nontrivial solution, linearly dependent/independent, linear transformation, oneto-one/injective, onto/surjective, diagonal matrix, upper/lower triangular matrix, matrix transpose, matrix
inverse, vector space, subspace, basis, null space/kernel, range, column space, row space, dimension, rank,
dimension, eigenvalue, eigenvector, eigenspace, characteristic polynomial, algebraic multiplicity, geometric
multiplicity, similar matrices
The exam will address some or all of the Learning Objectives listed above. The questions that follow are illustrations of the Learning Objectives, and not necessarily indicative of exam questions. Any formulas or definitions
needed to work through the review should be memorized for the exam.
Note that many of the questions below come from previous exam reviews.
1. Consider the linear system:

 −x1 + 3x2 = 0
x1 + 9x4
= 0

6x2 + x3
= 0
(a) Find the augmented coefficient matrix for the system.
(b) Find an REF and the RREF for the augmented matrix.
(c) In the augmented matrix, identify the pivot positions and pivot columns.
(d) Give the solution to the system.


−1 3 0 0
2. Let A =  1 0 0 9.
0 6 1 0
 
3
1

(a) Compute the product Au where u = 
3.
2
 
0
(b) Find all solutions to Ax = 21. (Hint: use previous questions to do this without reducing a new
9
matrix.)
 
 
 
 
−1
3
0
0







3. Let a1 = 1 , a2 = 0 , a3 = 0 , a4 = 9.
0
6
1
0
(a) Are the vectors linearly independent or dependent? If they are dependent, find the linear dependence
relation (the nontrivial set of weights that produce the zero vector).
(b) Find 4 vectors that are in Span{a1 , a2 , a3 , a4 }.
 
0
(c) Is b = 21 in Span{a1 , a2 , a3 , a4 }? If so, find the linear combination(s) of a1 , a2 , a3 , a4 that produce
9
b.
(d) Is Span{a1 , a2 , a3 , a4 } = R3 ?
 


x1
−x
+
3x
1
2
x2 
  

4. Let T : R4 → R3 be given by T 
x3  = x1 + 9x4 .
6x2 + x3
x4
(a) Is T linear? Prove your answer.
(b) If T is linear, find the standard matrix for T .
(c) If T is linear, determine whether T is onto R3 and whether T is one-to-one. Prove your answers.
2 3 −7 1
3 5
4 10
5. Let A =
,B =
, and C =
. Compute the following expressions if possible. If
4 2 5 0
1 2
6 15
an expression is undefined, state why.
(a) 3A − C
(c) AB
(e) AT C
(g) B 3
(i) B −1
(b) 2B + 21 C
(d) BA
(f) AC T
(h) A−1
(j) C −1
6. Find the inverse of the following matrices, and use it to solve Ax = b for the given vector b.


 
2
1
−3
1 1 0
4
(a) A =
,b=



−4 −1
7
(b) A = −3 0 1 , b = −3
−2 4 1
7
−5
4
7. A linear transformation T : R → R satisfies T (e1 ) =
and T (e2 ) =
. Find the standard matrix of
3
2
the inverse transformation T −1 .
2
2
8. Find the determinant of the matrix below


1 1 0
A = −3 0 1
−2 4 1
9. In each of the following problems, H is a subset of a vector space V . Determine and justify whether H is a
subspace of V . You do not need to prove that V is a vector space. If H is a subspace, find a basis for H and
find the dimension of H.
(a) V = P7 is the vector space of real polynomials of degree 7 or less; H is the subset of all polynomials in
P7 with a leading coefficient of 1 (these are called monic polynomials).
5t
(b) V = R2 ; H is the subset of all vectors in R2 of the form
.
t/2
(c) V = R2 ; H is the set of all eigenvectors of a 2 × 2 matrix A corresponding to a given eigenvalue λ of
geometric multiplicity 1.
 
x
3
3

(d) V = R ; H is the subset of all vectors in R of the form y  where z = xy.
z
(e) V = M3×3 ; H is the set of all 3 × 3 symmetric matrices. (A matrix A is symmetric if AT = A.)
10. Determine whether each of the following transformations is linear. If so, find its kernel and range.
(a) T : M2×2 → M2×2 , T (A) = A + AT .
(b) T : P2 → P4 , T (p(t)) = (p(t))2
(c) T : P2 → P4 , T (p(t)) = t2 · p(t)
11. Determine whether each of the following sets is a basis for the given vector space.
     
3
−1 
 2
(a) S = 8 , 12 , −4 ; V = R3


6
7
3
(b) S
(c) S
(d) S
(e) S
     
1
5 
 −1





3 , 0 , 0 ; V = R3
=


0
1
2
= {t2 + t + 1, t + 1, 1}; V = P2
= {t2 + t + 1, t2 − t − 1, t + 1}; V = P2
3 0
1 0
1 0
1 2
=
,
,
,
; V = M2×2
0 1
0 −2
0 3
2 1
     
1
5 
 −1
12. Consider the basis B =  3  , 0 , 0 for R3 .


0
1
2
(a) Find the change-of-coordinates matrix PB .
 
3

(b) Find the vector x whose coordinates relative to B are given by [x]B = 2.
1
 
3

(c) Find the coordinates of y = 2 relative to B.
1

2

13. Find the null space and column space of A = 1
1
the rank of A and the dimension of the null space.

10 0 −2
5 0 −1 , and find a basis for each space. Then find
5 2 −13
Does the Rank Theorem hold for A?
14. For each of the following matrices, find the characteristic polynomial and all eigenvalues and their corresponding eigenspaces. What is the algebraic and geometric multiplicity of each eigenvalue?




−9 0 7
1 2 3 4
1 2
(a) A =
0 1 3 0
8 1
(c) A =  0 5 0 

(d) A = 
0 0 0 5
−14
0
12
1 −2
(b) A =
0 0 0 2
2 5
15. Determine whether each of the following statements is true or false. If true, give justification. If false, explain
why or give a correction to the statement that will make it true.
You should also review the T/F statements from the previous exams and reviews.
(a) The geometric multiplicity of an eigenvalue λ for a matrix A is the multiplicity of λ in the characteristic
polynomial of A.
(b) An n × n matrix A must have exactly n eigenvalues, counted with multiplicity. (That is, if any repeated
eigenvalues are counted multiple times in accordance with their algebraic multiplicity.)
(c) If λ is an eigenvalue of A, then λk is an eigenvalue of Ak .
(d) Suppose an n × n matrix A has eigenvalue λ and corresponding eigenvector x, and another n × n matrix
B has an eigenvalue µ with the same corresponding eigenvector x. Then λµ is an eigenvalue of the
matrix AB.
(e) If λ and µ are two eigenvalues of a matrix A, then λ + µ is an eigenvalue of A.
(f) If A is an n × n matrix such that Ax = b has a solution x for every b in Rn , then all eigenvalues of A
are nonzero.
Linear Algebra Final Exam Review Solutions
For most problems, the final solution is given without supporting work. Make sure you understand how to
justify the answers. Refer to examples done in class to see the work for similar problems, if needed.


−1 3 0 0 0
1. (a) A =  1 0 0 9 0
0 6 1 0 0




1 −3 0 0 0
1 0 0 9 0
(b) One possible REF of A is 0 3 0 9 0. The RREF is 0 1 0 3 0.
0 0 1 −18 0
0 0 1 −18 0
(c) The pivot positions of A are the a11 , a22 , and a33 positions. The pivot columns are columns 1, 2, and 3.
 
−9
−3

(d) x = t 
 18 , for any real number t.
1
 
0
2. (a) Au = 21
9
(b) Since we solved the homogeneous system in the previous problem and
established
 
 that u is a particular
3
−9
1 −3
  
solution to the nonhomogeneous system, the general solution is x = 
3 + t  18  for any real number
2
1
t.
3. (a) The RREF of the matrix A = a1 a2 a3 a4 is given above. Since the matrix has a non-pivot column
(column 4), the vectors are linearly dependent. The dependence relation is
(−9t)a1 + (−3t)a2 + (18t)a3 + (t)a4 = 0 for any real number t.
As a specific example, −9a1 − 3a2 + 18a3 + a4 = 0.
(b) Answers will vary. The vectors a1 , a2 , a3 , a4 themselves 
are inSpan{a1 , a2 , a3 , a4 }, as is 0 and any
−7

other linear combination such as a1 − 2a2 + 3a3 − 4a4 = −35.
−9
Alternatively, one could show that the vectors are linearly dependent by pointing out that 4 vectors in
R3 can never be linearly independent.
(c) Yes, one can reduce the augmented matrix A = a1 a2 a3 a4 b to see whether the matrix is
consistent, and to find the weights so that c1 a1 +2 a2 + c3 a3 + c4 a4 = b. The linear combinations
are given by (21 − 9t)a1 + (7 − 3t)a2 + (−33 + 18t)a3 + (t)a4 = b for any real number t, or by
(3 − 9t)a1 + (1 − 3t)a2 + (3 + 18t)a3 + (2 + t)a4 = b for any real number t (the former is derived by
reducing [A | b] and the latter is derived from problem 2(b) above).
(d) Yes, the reduced form of
the matrix A = a 1 a2 a3 a4 shows that there3 are no zero rows, so the
augmented matrix A = a1 a2 a3 a4 b will be consistent for any b in R . Therefore, every vector
in R3 is in Span{a1 , a2 , a3 , a4 }.
 
 
u1
v1
u2 
v2 
4

 
4. (a) Yes, T is linear: Let c, d be scalars and let u = 
u3  , v = v3  be vectors in R . We need to show
u4
v4
that for any c, d, u, v, we have T (cu + dv) = cT (u) + dT (v):




cu1 + dv1
−(cu1 + dv1 ) + 3(cu2 + dv2 )
cu2 + dv2 

 

T (cu + dv) = T 
cu3 + dv3  = (cu1 + dv1 ) + 9(cu4 + dv4 )
6(cu2 + dv2 ) + (cu3 + dv3 )
cu4 + dv4




−u1 + 3u2
−v1 + 3v2
= c  u1 + 9u4  + d  v1 + 9v4  = cT (u) + dT (v)
6u2 + u3
6v2 + v3


−1 3 0 0
(b) A =  1 0 0 9
0 6 1 0
(c) The columns of A, as shown previously, span R3 , so T (x) = Ax = b has a solution for any b in R3 .
Therefore, yes, T is onto R3 .
However, the RREF also shows that there is a free variable. So when a solution to T (x) = Ax = b
exists, there are in fact infinitely many solutions. Therefore, no, T is not one-to-one.
5.
(a) Undefined: the sizes of 3A and C are not the same.
8 15
1
(b) 2B + 2 C =
5 11.5
(c) Undefined: the number of columns in A and the number of rows in B are not the same.
26 19 4 3
(d) BA =
10 7 3 1


32 80
24 60

(e) AT C = 
2 5
4 10
(f) Undefined: the number of columns in A and the number of rows in C T are not the same.
67 120
3
(g) B =
24 43
(h) Undefined: A is not square.
2 −5
−1
(i) B =
−1 3
(j) Undefined: C is singular because det C = 0.
−1/2 −1/2
−2
−1
−1
6. (a) A =
, and x = A b =
2
1
1


 
4/3
1/3 −1/3
2
−1
−1



(b) A = −1/3 −1/3 1/3 , and x = A b = 2
4
2
−1
3
−5 4
7. The standard matrix of T is A =
. The standard matrix of T −1 is A−1 =
3
2
−1/11 2/11
.
3/22 5/22
1
−22
2 −4
=
−3 −5
8. det A = −3
9.
(a) H is not a subspace of P7 : The zero polynomial 0 = 0t7 + ... + 0t + 0 is not an element of H, since it
does not have a leading coefficient.
5
(b) H is a subspace of R2 : H = Span
, so H must be a subspace. One possible basis is B =
1/2
5
. The dimension of H is 1.
1/2
(c) H is a subspace of R2 : Define H = {x ∈ R2 : Ax = λx} (or H = {x ∈ R2 : (A − λI)x = 0} and
adjust the following computations accordingly).
0 ∈ H, since A0 = 0 and λ0 = 0, so Ax = λx.
Suppose x and y both belong to H, so Ax = λx and Ay = λy are both true. Then A(x + y) =
(Ax) + (Ay) = (λx) + (λy) = λ(x + y), so x + y ∈ H. Thus H is closed under vector addition.
Suppose x ∈ H, so Ax = λx is true, and c ∈ R. Then A(cx) = c(Ax) = c(λx) = λ(cx), so cx ∈ H.
Thus H is closed under scalar multiplication.
Therefore H is a subspace.
Since the eigenvalue λ has geometric multiplicity 1, the dimension of H is 1. A basis for H is given by
any eigenvector corresponding to λ.
 
x
(d) H is not a subspace of R2 : Let v belong to H, so v = y  for some real numbers x, y, z satisfying
z
 
cx
z = xy. Let c be a real scalar. Then cv = cy . Observe (cx)(cy) = c2 xy 6= cz (unless c = 0 or
cz
c = 1), so in general cv is not in H.
(e) H is a subspace of M3×3 : The matrix 03×3 is symmetric, so 03×3 belongs to H.
Suppose A and B belong to H, and consider A + B: (A + B)T = AT + B T = A + B (where the last
equality is true because A = AT and B = B T ), so A + B is symmetric and thus H is closed under
vector addition.
Suppose A belongs to H, and consider cA for any real number c: (cA)T = cAT = cA (where the last
equality is true because A = AT ), so cA is symmetric and thus H is closed under scalar multiplication.
Therefore H is a subspace of M3×3 .


a b c
An arbitrary symmetric matrix will have the form A =  b d e  for real numbers a, b, c, d, e, f . One
c e f
possible basis is therefore

 
 
 
 
 

0 1 0
0 0 1
0 0 0
0 0 0
0 0 0 
 1 0 0
B = 0 0 0 , 1 0 0 , 0 0 0 , 0 1 0 , 0 0 1 , 0 0 0 .


0 0 0
0 0 0
1 0 0
0 0 0
0 1 0
0 0 1
The dimension of H is 6.
10.
(a) T is linear: For all 2 × 2 matrices A, B, and for all real scalars c, d, we have T (cA + dB) = (cA + dB) +
(cA + dB)T = cA + dB + cAT + dB T = c(A + AT ) + d(B + B T ) = cT (A) + dT (B). Ker(T ) is the
set of all matrices satisfying AT = −A (these are called skew-symmetic or antisymmetric matrices).
The range of T is the set of all symmetric matrices.
(b) T is not linear: Let p(t) = t2 and let q(t) = t. Then T ((p + q)(t)) = (t2 + t)2 = t4 + 2t3 + t2 . On the
other hand, T (p(t)) + T (q(t)) = (t2 )2 + (t)2 = t4 + t2 . Thus T ((p + q)(t)) 6= T (p(t)) + T (q(t)).
(c) T is linear: For all polynomials p(t), q(t) in P2 and for all real scalars c, d, we have T ((cp + dq)(t)) =
t2 ((cp + dq)(t)) = t2 (cp(t) + dq(t)) = ct2 p(t) + dt2 q(t) = cT (p(t)) + dT (q(t)). Ker(T ) is the zero
polynomial. The range of T is the set of all polynomials of the form at4 + bt3 + ct2 .


2 3 −1
11. (a) No, S is not a basis: Let A = 8 12 −4 be the matrix whose columns are the vectors in S. The
6 7 3


1 0 4
RREF of A is 0 1 −3, indicating that the columns of A (and thus the vectors in S) are not linearly
0 0 0
independent and don’t span R3 .


−1 1 5
(b) Yes, S is a basis: Let A =  3 0 0 be the matrix whose columns are the vectors in S. The RREF
0 1 2
of A is I3 , so the columns of A (and thus the vectors in S) are linearly independent and span R3 .
(c) Yes, S is a basis: First, c1 (t2 +t+1)+c2 (t+1)+c3 (1) = 0 = 0t2 +0t+0 implies that c1 = c2 = c3 = 0,
so the polynomials in S are linearly independent. Second, for any polynomial at2 +bt+c in P2 , if c1 = a,
c2 = b − a, and c3 = c − b, then c1 (t2 + t + 1) + c2 (t + 1) + c3 (1) = at2 + bt + c, so the polynomials
in S span P2 .
(d) No, S is not a basis: c1 (t2 + t + 1) + c2 (t2 − t − 1) + c3 (t + 1) = 0 = 0t2 + 0t + 0 is equivalent to the
system

 c1 + c2 = 0
c1 − c2 + c3 = 0

c1 − c2 + c3 = 0
which has solution c1 = − 21 s, c2 = 21 s, c3 = s for any real number s. Since the homogeneous equation
has a nontrivial solution, the vectors in S are not linearly independent.
1 0
1 0
1 2
3 0
0 0
(e) Yes, S is a basis: First, c1
+ c2
+ c3
+ c4
= 0 =
implies
0 1
0 −2
0 3
2 1
0 0
that c1 = c2 = c3 = c4 = 0, so the matrices in S are linearly independent. Second, for any matrix
a b
in M2×2 , if c1 = 31 (2a − 52 b − 32 c + d), c2 = 31 (a + b − d), c3 = 21 b, and c4 = 12 c, then
c d 1 0
1 0
1 2
3 0
a b
c1
+ c2
+ c3
+ c4
=
, so the matrices in S span M2×2 .
0 1
0 −2
0 3
2 1
c d


−1 1 5
12. (a) PB =  3 0 0
0 1 2
 
4

(b) x = PB [x]B = 9
4



  

0
1/3
0
0
1/3
0
3
2/3
(c) PB−1 = −2/3 −2/9 5/3 , and so [y]B = PB−1 y = −2/3 −2/9 5/3  2 = −7/9
1/3
1/9 −1/3
1/3
1/9 −1/3 1
8/9
   




−5s + t
−5
1 


1 5 0 −1
   

 s 
1  0





13. RREF of A is 0 0 1 −6 . Nul A = 
, so a basis for the null space is   ,   .
6t 
0
6 



0 0 0 0


t
0
1
dim Nul A = 2.
      

10
0
−2 
 2
Col A = Span 1 ,  5  , 0 ,  −1  . Since the pivot columns of A are columns 1 and 3, we could


1
5   2  
−13

   
0 
0 
 2
 2







1 , 0
1 , 0 . rank A = 2.
also write Col A = Span
. A basis for Col A is




1
2
1
2
The Rank Theorem says that if A is m × n, then rank A + dim Nul A = n. Indeed, 2 + 2 = 4.
1
1
2
14. (a) pA (λ) = λ − 2λ − 15. λ1 = −3, eigenspace Span
; λ2 = 5, eigenspace Span
. Both
−2
2
eigenvalues have algebraic and geometric multiplicity 1.
1
2
(b) pA (λ) = λ − 6λ + 9. λ = 3, eigenspace Span
. The eigenvalue has algebraic multiplicity 2
−1
and geometric multiplicity 1.
   
0 
 1
2



0 , 1 and algebraic and geometric
(c) pA (λ) = −1(λ − 5) (λ + 2). λ1 = 5 has eigenspace Span


2
0
 
 1 
multiplicity 2; λ2 = −2 has eigenspace Span 0 and algebraic and geometric multiplicity 1.


1
(d) pA (λ) = (1 − λ)2 (0 − λ)(2 − λ). The eigenvalues are λ = 1 (algebraic multiplicity2),
 λ= 0
1 



 
0
(algebraic multiplicity 1), λ = 2 (algebraic multiplicity 1). The eigenspaces are Span 
0 for





0
  
 
3 
53 




  

 

−3
 for λ = 0, and Span 15 for λ = 2. All eigenvalues have geometric
λ = 1, Span 
 1 
 5 










0
2
multiplicity 1.
15.
(a) False - The algebraic multiplicity of λ is the multiplicity of λ in the characteristic polynomial of A.
(b) True - The characteristic polynomial of A has degree n, so it must have exactly n roots, although they
may be complex or repeated.
(c) True - If Ax = λx, then Ak x = λk x.
(d) True - we know Ax = λx, and Bx = µx. Then (AB)x = A(Bx) = A(µx) = µ(Ax) = µ(λx) = λµx.
(e) False - Consider any 2 × 2 matrix with two distinct, nonzero eigenvalues λ and µ. Then these are the
only eigenvalues of A, and λ + µ 6= λ and λ + µ 6= µ, so λ + µ is not an eigenvalue.
(f) True - By the Invertible Matrix Theorem, these statements are equivalent.