Linear Equations and Matrices: Algebra Homework Assignment Solutions

This document presents solutions to a series of algebra problems focused on solving systems of linear equations. The problems cover various methods, including the substitution method for solving two-variable systems, the elimination method for similar systems, and Gaussian elimination for solving three-variable systems. Furthermore, matrix operations, specifically row reduction to reduced row echelon form (RREF), are employed to solve systems of equations. One problem involves setting up and solving a system of equations to determine the cost of items. Another focuses on finding the speeds of two cars based on distance and time relationships. Each solution provides a step-by-step breakdown of the process, leading to the final answers. Desklib provides more solved assignments and study resources for students.

Question one
y=4 x−3 … … … . i
2 y+8 x=26 … … …i i
Substitute y in equation ii
2 ( 4 x−3 ) + 8 x=26
8 x−6+ 8 x =26
16 x=32
x=2
Substitute x=2 in equation i
y=4 ( 2 ) −3
y=8−3
y=5
Answer:
x = 2
y = 5
Question Two
x +4 y=−24 … … … .. i
5 x+ 3 y =−35 … … … ii
Multiply equation i by 5, to eliminate x subtract equation ii from i
5 x+ 20 y =−120
5 x+3 y =−35
17 y=−85
y=−5
Substitute y=−5 in equation i to obtain the value of x
x +4 y=−24
x +4 (−5 )=−24
x−20=−24
x=−24 +20
x=−4

Answer:
x = -4
y = -5
Question Three
x + y + z=2
x− y +2 z=−3
2 x+ y + z =0
Matrix
(1 1 1
1 −1 2
2 1 1| 2
−3
0 )
Making the highlighted element to be 0: −2 R1+ R3
(1 1 1
1 −1 2
0 −1 −1| 2
−3
−6 )
Making the highlighted element to be 0: −R1 + R2
( 1 1 1
0 −2 −1
0 −1 −1| 2
−5
−2 )
Making the highlighted element to be 0: R2−2 R3
( 1 1 1
0 −2 −1
0 0 1 | 2
−5
−1 )
Making the highlighted element to be 1: −1
2 R2
( 1 1 1
0 1 1
2
0 0 1 | 2
5
2
−1 )

Answer:
( 1 1 1
0 1 1
2
0 0 1 | 2
5
2
−1 )System in triangular form: _________________________
Rewriting the system using the Triangular form
x + y +z=2 … … ..i
y + z
2 =5
2 … … … ii
z=−1
Substitutez=−1, in ii, then in i respectively
y + −1
2 =5
2
y− 1
2 = 5
2
y=3
Substitute z=−1 and y=3 in equation ii
x +3−1=2
x=0
x = _______0_____
y = ______3______
z = _______-1_____
Question Four
x - y + 5z = -11
3x + z = -2
x + 4y + z = 2

(1 −1 5
3 0 1
1 4 1|−11
−2
2 )
Making the highlighted element to be 0: R1−R3
( 1 −1 5
3 0 1
0 −5 4 |
−11
−2
−13 )
Making the highlighted element to be 0: 3 R1−R2
(1 −1 5
0 −3 14
0 −5 4 |−11
−31
−13 )
Making the highlighted element to be 0: 5 R1−R3
(1 −1 5
0 −3 14
0 0 21|−11
−31
−42 )
Making the highlighted element to be 0: 3 R1−R 2
(1 0 1
0 −3 14
0 0 21| −2
−31
−42 )
Making the highlighted element to be 0: 14 R1−R2
(1 0 0
0 −3 14
0 0 21| 3
−31
−42 )
Making the highlighted element to be 0: 3
2 R2−R 3
(1 0 0
0 −3 0
0 0 21| 3
−9
2
−42 )

(1 0 0
0 −3 0
0 0 21| 3
−9
2
−42 )
Making the highlighted element to be 0: 1
3 R2 and 1
21 R3
Answer:
RREF Matrix:
(1 0 0
0 1 0
0 0 1| 3
−3
2
−2 )
x = ______3______
y = ____-3/2________
z = ________-2____
Question Five
2x + y + 2z - 4w = 10
x + 3y + 2z - 11w = 17
3x + y + 7z - 21w = 0
( 2
1
3
0
1
3
1
0
2
2
7
0
−4
−11
−21
0 |
10
17
0
0 )
Making the highlighted element to be 0: 3 R2−R3
(2
1
0
0
1
3
8
0
2
2
−1
0
−4
−11
−12
0 |10
17
51
0 )
Making the highlighted element to be 0: R1−2 R2

(2
0
0
0
1
−5
8
0
2
−2
−1
0
−4
18
−12
0 | 10
−24
51
0 )
Making the highlighted element to be 0: R2 + 5
8 R3
( 2
0
0
0
1
−5
0
0
2
−2
−21
8
0
−4
18
−19
2
0 | 10
−24
63
8
0 )
Making the highlighted element to be 0: 5 R1 + R2
(10
0
0
0
0
−5
0
0
8
−2
−21
8
0
−2
18
−19
2
0 | 26
−24
63
8
0 )
Making the highlighted element to be 0: R2 + 2304
325 R1
( 945
4
0
0
0
0
−20
0
0
0
0
−21
8
0
0
72
−19
2
0 | 525
32
−72
63
8
0
)
Making the highlighted element to be 1: 1
72 R2
and −2
19

( 945
4
0
0
0
0
−20
0
0
0
0
−21
8
0
0
1
1
0| 4437
4
−1
63
76
0
)Making the highlighted element s to be 1: 4
945 R1 , −1
20 R2 , −8
21 R3 , R3 +R4
(1
0
0
0
0
1
0
0
0
0
1
0
0
1
1
1
|4437
945
1
20
−12
13
63
76
)Answer:
RREF Matrix:
(1
0
0
0
0
1
0
0
0
0
1
0
0
1
1
1
|4437
945
1
20
−12
13
63
76
)
x = ____ __ 4437
945 ______
y = _______ 1
20 _____
z = ___________ −12
13 _
w = _________ 63
76 __
Question Six
Let t ,b∧s be the price of tent, sleeping bag and camp stool respectively

Equations:
t=9 s … … i
b=s +23 … … . ii
t+ 5 b+2 s=227 … .. iii
To obtain the value of the three variables, t ,b∧s substitute t=9 s and b=s +23 in equation iii
t+ 5 b+2 s=227
9 s+5 ( s+23 ) +2 s=227
9 s+5 s+115+2 s=227
16 s=227−115
16 s=112
s=7
Substitute s=7 in equation i and ii to obtain the value of t and b respectively.
t=9 s
t=9 ( 7 )
t=63
b=s +23
b=7+23
b=30
Answer:
Cost of a tent: _____63_____
Cost of sleeping bag: __30________
Cost of a camp stool: __7________
Question Seven
Let the speed of the faster car be x and that of slower car be y, and since the distance is given by
speed multiply by time then,

3 x−3 y=24 … … i
Since, time is given by distance divided by time, then time for the slower car:
159
y =3 … ..ii
Therefore, the two equations are:
3 x−3 y=24 … … i
159
y =3 … ..ii
Solve by substitution;
159
y =3 … ..ii
3 y=159
y=53
Substitute y=53 in equation ito obtain the value of x
3 x−3 ( 53 )=24
3 x−159=24
3 x=24+ 159
3 x=183
x=61
Answer:
Speed of Faster car: _____61_______ mph; Speed of Slower car: _______53_______ mph.