University Math 4A, Winter 2020: Linear Algebra Homework Assignment 1

This document presents a comprehensive solution to a Math 4A homework assignment focused on linear algebra concepts. The assignment covers solving systems of linear equations using elimination, determining if points lie on lines, and understanding echelon forms. It includes problems on identifying consistent and inconsistent systems, determining the number of solutions (unique, infinite, or none), and applying these concepts to various linear systems. The solution demonstrates the use of row operations, the analysis of augmented matrices, and the identification of pivot and free variables. Furthermore, it addresses concepts such as the consistency of systems based on determinant values and the application of row operations. The assignment also includes true/false questions about linear algebra principles and matrix transformations. The final part of the solution involves identifying the form of matrices (echelon and reduced echelon form).

Bushra Alrashdi
Assignment HW1
1. Solve the system using elimination
6x−3y=9 5x−3y=6
6x-3y=9 ……….(1)
5x-3y=6………..(2)
(1)×5-(2) × 6 gives
30x-15y=45……….(3)
30x-18y=36…………(4)
---------------
-3y=-9
---------------
y=3
substituting the value of “y” in (3), we get,
30x-15(3)=45
30x=45+45=90
x=3
therefore, x=y=3
2. Determine which of the points (−2,−4), (4,−5), and (−1,−1)lie on both the lines −4x1−5x2
= 9 and 3x1−x2 = −2.
Here we use trial and error method to find the points which lie on both the given lines.
Condition for points to be on line is that, they should satisfy the equation of lines.
Let us check for (-1,-1)
Given: -4x1-5x2=9
-4(-1)-5(-1) = 4+5=9
Given: 3x1-x2=-2
3(-1)-(-1)=-2

Therefore (-1,-1) lie on both the given lines.
Row echelon form:
The leading (first) entry in each row must be equal to one.
The leading entry on each subsequent row must be on a new column to the right.
All rows where all entries are zero are below rows where NOT all entries are zero.
3. Determine if the linear system −x1 − 2x2 = −2, − 7x2 = 0 is in echelon form
C. The system is not in echelon form because not every equation has a leading variable.
4. A linear system with three equations and two variables must be inconsistent
False.
Condition for system solutions:
If a1/a2 = b1/b2 = c1/c2 then, infinite solutions
If a1/a2 = b1/b2 ≠ c1/c2 then, no solution
If a1/a2 ≠ b1/b2, then, unique solution
5. For the following linear systems enter the letter U if the system has a unique solution, the
letter N if it has no solution, and the letter I if it has infinitely many solutions.
3x + 4y = 1, 3x − 4y = 1
Since, a1/a2 ≠ b1/b2, the system has unique solution
3x + 4y = 1, 6x + 8y = 1
Since, a1/a2 = b1/b2 ≠ c1/c2, the system has no solution
3x + 4y = 1, 6x + 8y = 2
Since, a1/a2 = b1/b2 = c1/c2, the system has infinitely many solutions.
6. 12x − 15y = 15, 20x − 25y = k
For the above system of equations to be consistent, k must equal

For the above system of equations to be consistent, the condition is,
a1/a2 = b1/b2 = c1/c2
Therefore, we have,
12/20 = -15/-25 = 15/k
3/5 = 3/5 = 15/k
3/5 = 15/k
K=(15*5)/3 = 25
Therefore, k=25
7. Give a geometric description of the following systems of equations
1. −x − 3y = −4 6x + 8y = −8 ?
Two lines intersecting at a point.
2. 4x + 4y = −2 10x + 10y = −5 ?
Two lines that are same.
3. 4x + 4y = −2 10x + 10y = −8
Two parallel lines.
8. Solve the following system of equations. If there are no solutions, type ”No Solution” for
both x and y. If there are infinitely many solutions, type ”x” for x, and an expression in
terms of x for y. 1x+2y = 2, −1x−2y = −2
Since, for a1/a2 = b1/b2 = c1/c2, infinite solutions
From given, we have,
1/-1 = 2/-2 = 2/-2
-1/1 = -1/1 = -1/1
Therefore, the given system of equations, has infinite solutions
x=x
2y=2-x
9. Determine which of the points (5,−4,−4), (6,−2,−1), and (1,−3,−2) satisfy the linear
system
2x1 − 5x2 − 2x3 = 21, −2x1 + 8x2 + 3x3 = −32
Here we use trial and error method to find the points which satisfy given system.
Condition for points to satisfy the system is that, they should satisfy the equation of lines.
Let us check for (1,-3,-2)

Given: 2x1 − 5x2 − 2x3 = 21
2(1) – 5(-3) – 2(-2) = 21
Given: −2x1 + 8x2 + 3x3 = −32
−2(1) + 8(-3) + 3(-2) = −32
Therefore (1,-3,-2) satisfy linear system.
10. Determine if the linear system x1 − x2 − x3 = 0, 2x3 = −4, 2x2 + 2x3 = 13 is in echelon
form
The system is not in echelon form because not every equation has a leading variable.
11. Find the set of solutions for the linear system x1 + 2x2 + 6x3 = 1, − x2 + 2x3 = 8, 5x3 = 5
5x3 = 5  x3=1
− x2 + 2x3 = 8  -x2+2(1) =8  -x2=8-2=6  x2=-6
x1 + 2x2 + 6x3 = 1  x1+2(-6)+6(1) =1  x1=7
12. Find the set of solutions for the linear system −2x1 + 5x2 + 3x3 = 11, − 6x2 − 9x3 = −7
x3 = x3
− 6x2 − 9x3 = −7  x2 = - (-7+9x3) / 6
−2x1 + 5x2 + 3x3 = 11 −2x1 + 5 [- (-7+9x3) / 6 ] + 3x3 = 11  x1 = - (27x3+31) /
12
If x3=0, then,
x2 = 7/6
x1 = - 31/12
13. For the following system to be consistent, k ≠
-3x-6y-6z=7

-17x-6y+kz=-8
2x-3y+8z=8
For the system to be consistent Δ ≠ 0
-3 -6 -6
-17 -6 k
2 -3 8
-3(-48+3k) +6(-136-2k) -6(63) ≠ 0
-21k-1050 ≠ 0
k ≠ -50
14. Solve the system using row operations (or elementary
matrices).
-6x-4y-3z=-17
-3x+6y+2z=11
-6x+3y+6z=6
R3=R3-R1
R2=R2-R1
-6x-4y-3z=-17
0+10y+5z=28
0+7y+9z=23
Solving 2nd and 3rd equations, we get,
Z=1, y=2
Substituting the value of y and z in 1st equation, we get,
X=1
z=1, y=2, x=1
15.
16. Row reduced matrix form of
x-5y-3z=-5
-x+2y+15z=14
-x+3y+11z=11
R2=R2+R1
R3=R3+R1

R2=R2 / -3
R1=R1+5R2
R3=R3+2R2
The reduced form is,
1 0 -23 -20
0 1 -4 -3
0 0 0 0
17. Given the augmented matrix below,
1 -2 5 8 6 3 | -5
0 0 0 1 8 9 | 9
0 0 0 0 1 -9 | -5
We have,
x5-9x6 = -5
x4+8x5+8x6 = 9
x1-2x2+5x3+8x4+6x5+3x6 = -5
18. A Suppose that the echelon form of an augmented matrix has a pivot position in every
column except the rightmost one. How many solutions does the associated linear system
have?
A linear system is consistent if and only if the rightmost column of the augment matrix is
not a pivot column. Since the system is consistent, it has one and only one solution.
Exactly one solution.
19. Determine how many pivots each of the following matrices have
(1) 1 0 0 0 10 0 1 0 0 −9 0 0 1 0 −3 0 0 0 1 7       
1st element of each row
4 pivot values.
(2) 1 0 9 0 0 0 0 0 0   
1st element of first row
1 pivot value.
(3) 1 0 −2 8 0 1 −10 −6 0 0 0 0 0 0 0 0       
1st element of each row
2 pivot values

(4) 0 1 0 6 0 0 1 6
2 pivot values
20. How many free variables does each augmented matrix have?
(1) 1 0 4 0 1 2 0 0 0   
None
(2) 1 0 0 2 −6 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0       
One (x4 = 2)
(3) 1 −8 2 0 0 0 0 0 0   
One (x2=-8)
(4) 1 −9 9 −5 0 0 0 0
Two, (x2=-9, x3=9)
21. How many determined (basic) variables does each augmented matrix have?
(1) 1 0 0 −8 10 0 1 0 0 −7 0 0 1 0 9   
3 basic variables, 1st element of each row consisting of 1
(2) 1 6 −5 0 0 0 0 0 0   
1 basic variable, element of 1st row
(3) 1 0 −6 0 1 −8 0 0 0   
2 basic variables, 1 st element of each row consisting of 1
(4) 1 0 0 10 −5 0 1 3 0 −8 0 0 0 0 0   
2 basic variables, 1st element of each row consisting of 1.
22. The reduced row-echelon forms of the augmented matrices of four systems are given
below. How many solutions does each system have?
(1) 1 0 0 17 0 0 1 −11
No solution
(2) 1 0 5 0 1 −3
Unique solution
(3) 1 0 −2 0 0 1 0 0 0 0 0 1 0 0 0 0       

Infinitely many solutions
(4) 1 0 10 0 1 19 0 0 0   
Unique solution
23. Determine the value of h such that the matrix is the augmented matrix of a consistent
linear system.
7 -6 | h
-14 12 | 5
Perform row operation
R2 = 2R1+R2
7 -6 | h
0 0 | 2h+5
2h+5=0
h = -5/2
24. Consider a linear system whose augmented matrix is
1 1 4 | -2
1 2 -4 | 0
6 14 k | 5
For the system to have no solution, determinant should be zero.
Upon performing the determinant of matrix, we get.
K+40=0
k=-40
25.
1. If the solution to a system of linear equations is given by (4 - 2z, -3 + z, z), then (4, -
3, 0) is a solution to the system.
TRUE
2. If the entries of the bottom row of an augmented matrix in reduced row-echelon form
are 0, 1, 3, 1, then the system has no solution.
FALSE
3. If a matrix is in reduced-row echelon form, then the first nonzero entry in each row is
a 1 and has 0’s below it.
TRUE

26. The matrix on the right results after performing a single row operation on the matrix on
the left.
Identify the row operation. Use R1, R2, and R3 to represent the rows R1, R2, and R3. For
example, if the row operation were ”add 2 times row 4 to row 6” then the answer would
be ”2 R4 + R6 −→ R6”. If it were ”swap row 4 with row 6” the answer would be ”R4∗
←→ R6”
R2 = -2 *R2
27. The matrix on the right results after performing a single row operation on the matrix on
the left.
Identify the row operation. Use R1, R2, and R3 to represent the rows R1, R2, and R3. For
example, if the row operation were ”add 2 times row 4 to row 6” then the answer would
be ”2 R4 + R6 −→ R6”. If it were ”swap row 4 with row 6” the answer would be ”R4∗
←→ R6”
R3=R3-2R2
28. If two matrices are equivalent, then one can be transformed into the other with a sequence
of elementary row operations.
True
29. Different sequences of row operations can lead to different echelon forms for the same
matrix.
True
30. Different sequences of row operations can lead to different reduced echelon forms for the
same matrix
False
31. Given the augmented matrix
A = 1 −1 −4 −1 2 −1 1 3 −2 −2 −6 1 ,   
perform each row operation in the order specified and enter the final result.
First: R2 −2R1 → R2,
1 -1 -4 | -1
0 1 9 | 5
-2 -2 - 6 | 1
Second: R3 +2R1 → R3,
1 -1 -4 | -1

1 1 9 | 5
0 -4 -14 | -1
Third: R3 +4R2 → R3.
1 -1 -4 | -1
2 1 9 | 5
6 -6 -2 | 13
32. Determine whether the following matrices are in echelon form, reduced echelon form or
not in echelon form.
(1) 0 1 0 −9 0 0 1 7
Echelon form
(2) 9 1 1 0 0 1 −10 0 0 0 0 10   
Row reduced echelon form
(3) 7 1 1 1 −7 3 0 0 1   
Not in echelon form
(4) 1 0 1 0 1 1 0 0 −6   
Reduced echelon form
33.
Row echelon form:
The leading (first) entry in each row must be equal to one.
The leading entry on each subsequent row must be on a new column to the right.
All rows where all entries are zero are below rows where NOT all entries are zero.
Reduced echelon form:
the leading entry is the only nonzero entry in its column.
(1) 1 1 7 1 0 −8
Not in echelon form
(2) 8 −6 −9 −4 −3 0 −10 1 1 1 0 0 1 0 4 0 0 0 1 0       
Reduced echelon form
(3) 1 0 0 −3 0 1 0 −3 0 0 0 0   
Echelon form

(4) 7 0 1 0 5 0
Reduced echelon form
34. Reduce the matrix
A = 3 −2 −2 11 3 2 −2 −1 −1 1 2 −2   
A = 3 -2 -2 11
3 2 -2 -1
-1 1 2 -2
Perform the below row operations, to get reduced row matrix form
R2 = R2-R1
R1 = R1/3
R3 = R3+R1
R2 = R2/4
R1 = R1+2/3 R2
R3 = R3 -1/3 R1
R3 = ¾ R3
R1 = R1+2/3R3
Final reduced row echelon matrix
1 0 0 3
0 1 0 -3
0 0 1 2
35. Let A = 1 0 −3 0 2 0 1 −3 0 1 0 0 0 1 −4 0 0 0 0 0       
Is the matrix in echelon form? (input Yes or No) Is the matrix in reduced echelon form?
(input Yes or No) If this matrix were the augmented matrix for a system of linear
equations, would the system be inconsistent, dependent, or independent?
The matrix is not in echelon form.
The matrix is in reduced echelon form.
A = 1 0 -3 0 | 2
0 1 -3 0 | 1
0 0 0 1 | -4
0 0 0 0 | 0
x1-3x3 = 2
x2-3x2 = 1
x4=-4

if x3=0, x2=1
x1=2
Therefore, the system is dependent. (on x3)