Fall 2022 Habib University Linear Algebra

Exercise Set 1.1 Solutions

Question 01:

What is a linear equation?

(a) It’s linear because all the variables 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1 and it is equal to a constant.
(b) It’s not linear because 𝑥1 𝑥3 are two different variables multiplying which each other and it does
not follow the above format of a linear equation. Despite the fact that both individual variables
have a degree of 1, they are multiplying with each other instead of a constant coefficient.
(c) It’s linear because all the variables 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1 and it is equal to a constant.
(d) It’s not linear because not all the variables 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1. The degree of 𝑥1 is
−2. It is not a positive integer.
(e) It’s not linear because not all the variables 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1. The degree of 𝑥1 is
3
. It is not a positive integer.
5
(f) Its linear because all the variables 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1 and it is equal to a constant.

Question 02:

(a) Since sin(𝑘) = 𝐶, where 𝐶 is a constant.

𝑥1 − 𝑥2 + 𝑥3 = 𝐶
It’s a linear equation because all the 𝑥1 , 𝑥2 and, 𝑥3 have a degree of 1 and it is equal to a
constant.
(b) Rewriting the given equation in a simplified format by removing the fraction element.
1
𝑘𝑥1 − 𝑥2 = 9 → 𝑘 2 𝑥1 − 𝑥2 = 9𝑘
𝑘
As 𝑘 is a constant, then 𝑘 2 and 9𝑘 are constants as well. Since all the given variables have a
degree of 1 and it is equal to a constant value of 9𝑘, the given equation is linear.
(c) 2𝑘 = 𝐶, where 𝐶 is a constant. Therefore, all the variables in the given equation have a degree
of 1 and it is equal to a constant value, the equation is linear.

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Fall 2022 Habib University Linear Algebra

Question 03:

A free variable is a variable used to represent a numerical substitution and is not a parameter itself or
part of the equation, it is used to specify a place in an expression where we can substitute numbers to
obtain results and helps us obtain a general form of our expression. In the solutions below we chose one
variable in each equation to be our output and then made it the subject of our equation and expressed
it in terms of other variables in the equation and finally replaced those variables with free variables to
obtain a general solution set. The solutions below are one possibility, we can use any variable to be our
output variable and then express it in terms of other variables.
(a)
7𝑥 − 5𝑦 = 3
7 3
𝑦= 𝑥−
5 5
7 3
𝑦 = 5 𝑡 − 5 , where 𝑥 = 𝑡 is a free variable
7 3
Solution Set = {(𝑥, 𝑦): 𝑦 = 5 𝑥 − 5}
7 3
= {(𝑥, 𝑥 − ) : 𝑥 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑖𝑠 𝑎 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
5 5
7 3
= {(𝑡, 𝑡 − ) : 𝑡 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑡 𝑖𝑠 𝑎 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
5 5

(b)
3𝑥1 − 5𝑥2 + 4𝑥3 = 7
7 5 3
𝑥3 = + 𝑥2 − 𝑥1
4 4 4
7 5 3
𝑥3 = 4 + 4 𝑠 − 4 𝑡, where 𝑥1 = 𝑡, 𝑥2 = 𝑠 are free variables
7 5 3
Solution Set = {(𝑥1 , 𝑥2 , 𝑥3 ): 𝑥3 = 4 + 4 𝑥2 − 4 𝑥1 }
7 5 3
= {(𝑥1 , 𝑥2 , + 𝑥2 − 𝑥1 ) : 𝑥1 , 𝑥2 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑥1 𝑎𝑛𝑑 𝑥2 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
4 4 4
7 5 3
= {(𝑡, 𝑠, + 𝑠 − 𝑡) : 𝑠, 𝑡 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑠 𝑎𝑛𝑑 𝑡 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
4 4 4
(c)
−8𝑥1 + 2𝑥2 − 5𝑥3 + 6𝑥4 = 1
1 5 2 8
𝑥4 = + 𝑥3 − 𝑥2 + 𝑥1
6 6 6 6
1 5 2 8
𝑥4 = + 𝑢 − 𝑡 + 𝑠, where 𝑥1 = 𝑠, 𝑥2 = 𝑡, 𝑥3 = 𝑢 are free variables
6 6 6 6
1 5 2 8
Solution Set = {(𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ): 𝑥4 = 6 + 6 𝑥3 − 6 𝑥2 + 6 𝑥1 }
1 5 2 8
= {(𝑥1 , 𝑥2 , 𝑥3 , + 𝑥3 − 𝑥2 + 𝑥1 ) : 𝑥1 , 𝑥2 , 𝑥3
6 6 6 6
∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑥1 , 𝑥2 𝑎𝑛𝑑 𝑥3 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
1 5 2 8
= {(𝑠, 𝑡, 𝑢, + 𝑢 − 𝑡 + 𝑠) : 𝑠, 𝑡, 𝑢 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑠, 𝑡 𝑎𝑛𝑑 𝑢 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
6 6 6 6

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Fall 2022 Habib University Linear Algebra

(d)
3𝑣 − 8𝑤 + 2𝑥 − 𝑦 + 4𝑧 = 0
1 2 8 3
𝑧= 𝑦− 𝑥+ 𝑤− 𝑣
4 4 4 4
1 2 8 3
𝑧 = 𝑑 − 𝑐 + 𝑏 − 𝑎, where 𝑣 = 𝑎, 𝑤 = 𝑏, 𝑥 = 𝑐, 𝑦 = 𝑑 are free variables
4 4 4 4
1 5 2 8
Solution Set = {(𝑣, 𝑤, 𝑥, 𝑦): 𝑥4 = + 𝑥3 − 𝑥2 + 𝑥1 }
6 6 6 6
1 2 8 3
= {(𝑣, 𝑤, 𝑥, 𝑦, 𝑦 − 𝑥 + 𝑤 − 𝑣) : 𝑣, 𝑤, 𝑥, 𝑦 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑣, 𝑤, 𝑥 𝑎𝑛𝑑 𝑦 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
4 4 4 4
1 2 8 3
= {(𝑎, 𝑏, 𝑐, 𝑑, 𝑑 − 𝑐 + 𝑏 − 𝑎) : 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝑅} 𝑤ℎ𝑒𝑟𝑒 𝑎, 𝑏, 𝑐 𝑎𝑛𝑑 𝑑 𝑎𝑟𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
4 4 4 4

Question 04:

What is an augmented matrix?

(c)

𝑥1 + 2𝑥2 − 𝑥4 + 𝑥5 = 1
3𝑥2 + 𝑥3 − 𝑥5 = 2
𝑥3 + 7𝑥4 = 1
Augmented Matrix for this system of linear equations is,
1 2 0 −1 1 1
(𝐴|𝐵) = [0 3 1 0 −1|7]
0 0 1 7 0 0

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Fall 2022 Habib University Linear Algebra

(d)
𝑥1 = 1
𝑥2 = 2
𝑥3 = 3
Augmented Matrix for this system of linear equations is,
1 0 01
(𝐴|𝐵) = [0 1 0|2]
0 0 13
Question 05:

(c)
7 2 1 −3 5
(𝐴|𝐵) = [ | ]
1 2 4 0 1
Linear system of equations for above augmented matrix is,

7𝑥1 + 2𝑥2 + 𝑥3 − 3𝑥4 = 5

𝑥1 + 2𝑥2 + 4𝑥3 = 1
Question 07:

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Fall 2022 Habib University Linear Algebra

Question 08:

pg. 5
Fall 2022 Habib University Linear Algebra

Question 09:

pg. 6