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UCLA MATH 33A (MATH33A) Weeks 1–9 Homework Bundle | Complete Questions & Solutions | All in 1 PDF | Spring 2026/2027.

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Complete UCLA MATH 33A (MATH33A) – Weeks 1–9 Homework Bundle, featuring 100% complete questions and worked solutions updated for Spring 2026/2027. Includes all homework from Weeks 1 through 9 bundled in one PDF file in a clear, organized format for easy reference. Ideal for students searching for UCLA MATH 33A homework solutions, MATH33A Weeks 1–9 answers, Linear Algebra homework solutions, and UCLA MATH 33A Spring 2026/2027 course content.

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UCLA MATH 33A (MATH33A) Weeks 1–9 Homework Bundle |
Complete Questions & Solutions | All in 1 PDF | Spring 2026/2027
Math 33A (Spring ’26) Put Your Name In This Space!
Week 1 Homework


1. (Discussed by Mon, Week 1 and Wed, Week 1) For all of the systems of equations below, find
all solutions using the reduced row echelon form method1 . Make sure to show your work.

y + 3x + 2z = 0

(a) −x + 5y + z = −3

2x − 7y + z = −6



 a+b+c+d=2

b + 7c + d = −15
(b)


 2a + 2b − 8c = 26

3b + 3c + 3d = 9

−y + x + z = 7

(c) x + y + z = 8

5x + 3y + 5z = 47


2x5 + x4 − x6 = 2

(d) x1 + 2x2 + x5 − x6 = 0

x1 + 2x2 + 2x3 − x5 + x6 = 2

 
x1
   x2   
1 8 16 −4 −5   x3  = 6
(e)

0 0 1 −7 18    1
x4 
x5
2. (Discussed by Wed, Week 1) (This one is slightly tricky!) Consider the following system of
linear equations, where k is an unknown constant:

x + y + kz =2
3x + 4y + 2z =k
2x + 3y − z =1

Find all values of k for which the system has:

(a) No solution
(b) A unique solution
(c) Infinitely many solutions
1
While it is okay to use any non-computer/calculator method, essentially the only method that can guarantee
that you have found all solutions is the reduced row echelon form method we discussed in class!

, Make sure to show your work using the augmented matrix and row reduction.

3. (Discussed by Wed, Week 1) Evaluate the following vector additions, or state that they are
not defined and explain why briefly.
   
8 6
1 4
(a) 7 + 5
  

2 0
   
2.5 0.9
(b) +
3.3 1.2
 
3  
2 2
(c) 
−1 + 3


0
   
2 −2
(d) −1 +  1 
−5 5
4. (Discussed by Wed, Week 1) Evaluate the following products and leave no arithmetic left to
be done, or state that they are not defined and explain why briefly.
  
8 1 2
(a)
7 2 −2
  
2 1 3 1
(b)
3 1 2 1
  
3 1 2 1
(c)
2 1 3 1
  
2 1 2 2 1
(d)
3 1 −1 3 1
 
 x
1 0 0  1

(e) x2 where x1 , x2 , x3 are arbitrary scalars.
0 1 0
x3
 
  −4 11
4 5 6
5. (Discussed by Wed, Week 1) Evaluate the product  61 100  and leave
101 4 −70
12 −70
no arithmetic left to be done.

6. (Discussed by Fri, Week 1)
 
0 1
(a) Define a function T : R → R given by the formula T (⃗v ) :=
2 2
⃗v . What
−1 0



2

, familiar geometric operation does this function correspond to?2 It is enough to give an
answer only, no work needs to be shown.
 
1 0
(b) Define a function R : R2 → R2 given by the formula R(⃗v ) := ⃗v . What
0 −1
familiar geometric operation does this function correspond to? It is enough to give an
answer only, no work needs to be shown.
   
x x
(c) There is a 2 × 2 matrix M such that, for any vector ∈R ,M
2
is the same
  y y
x
as the vector you get starting from , then rotating the plane 90 degrees counter-
y
clockwise, and then reflecting over the y-axis (so that the y-axis is fixed). What is M ?
On this problem some work must be shown.




2
Hint: You may want to look at part (c) for language you can use to express your answer.

3

, Solutions
1. (a) First, make sure that you write the first equation in ‘alphabetical order’, meaning you
write
 it as 3x + y + 2z  = 0. Once you do that, you can write the augmented matrix
3 1 2 0
as −1 5 1 −3.
2 −7 1 −6
Let us begin the reduction steps by swapping the first and the second rows since we’ll
get a coefficient of −1 at the top left.
     
3 1 2 0 −1 5 1 −3 R1 1 −5 −1 3
Swap R1 and R2 −1
−1 5 1 −3 − −−−−−−−→  3 1 2 0  −−→ 3 1 2 0
2 −7 1 −6 2 −7 1 −6 2 −7 1 −6
   
1 −5 −1 3 1 −5 −1 3
R2 +(−3)R1 R3 +(−2)R1
−−−−−−→ 0 16 5 −9 −−−−−−→ 0 16 5 −9 
2 −7 1 −6 0 3 3 −12
   
1 −5 −1 3 R2 1 −5 −1 3
Swap R2 and R3 3
−−−−−−−−→ 0 3 3 −12 −−→ 0 1 1 −4
0 16 5 −9 0 16 5 −9
   
1 −5 −1 3 R3 1 −5 −1 3
R3 +(−16)R2 −11
−−−−−−−→ 0 1 1 −4 −−→ 0 1 1 −4
0 0 −11 55 0 0 1 −5
   
1 −5 −1 3 1 −5 0 −2
R2 +(−1)R3 R1 +(1)R3
−−−−−−→ 0 1 0 1  −−−−−→ 0 1 0 1 
0 0 1 −5 0 0 1 −5
 
1 0 0 3
R1 +(5)R2
−−−−−→ 0 1 0 1 

0 0 1 −5

All these steps do not change the solutions of the system. The last matrix is in reduced
row echelon form and so, we can see from that the only solution of the system is
x = 3, y = 1, z = −5.
(In one of the steps, we swapped R2 and R3 just to make our calculations easier. You
could have proceeded without swapping as well, in that case you would have arrived
at the same answer but the steps would have been more tedious.)
 
1 1 1 1 2
0 1 7 1 −15
(b) The augmented matrix is  2 2 −8 0 26 .


0 3 3 3 9




4

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