Linear Algebra Assignment: Matrix Operations and Gaussian Elimination

This document presents a detailed solution to a linear algebra assignment. The assignment covers key concepts such as expressing vectors as linear combinations, solving matrix equations, and performing matrix operations. The solution includes step-by-step explanations and computations, demonstrating how to solve for unknown variables in matrix equations. Furthermore, the assignment explores Gaussian elimination to solve systems of linear equations represented in matrix form. The solution provides the computation of 3x3 matrices using matrix operations and then utilizes Gaussian elimination and back substitution to find the solution. The assignment offers a comprehensive guide to understanding and solving linear algebra problems, making it a valuable resource for students studying the subject.

3/22/2019 Math Assignment
Linear Algebra
[Student Name]
[STUDENT ID]

Ques 1. Express 𝒃 = [
−𝟏
−𝟐
𝟕
] as a linear combination of 𝒗𝟏 = [
𝟐
𝟒
𝟒
], 𝒗𝟐 = [
−𝟑
−𝟔
𝟑
] , and 𝒗𝟑 =
[
𝟒
𝟏𝟎
𝟏𝟎
] . Express the problem as a matrix equation.
Sol. it is clear that 𝑏 can be easily calculated by 𝑏 = 𝑣1 + 𝑣2
[
−1
−2
7
] = [
2
4
4
] + [
−3
−6
3
]
Ques 2. a) Find any numbers 𝒚and 𝒛such that
[
𝟏 𝟐
𝟐 𝟒
] [
𝒚
𝒛
] = [
𝟏
𝟐
]
Sol.
[
1 2
2 4
] [
𝑦
𝑧
] = [
1
2
]
[
1𝑦 + 2𝑧
2𝑦 + 4𝑧
] = [
1
2
]
Comparing LHS and RHS
𝑦 + 2𝑧 = 1 … … … . (1)
2𝑦 + 4𝑧 = 2 … … … (2)
From Eq (1) and (2) 𝑦 = 1 𝑎𝑛𝑑 𝑧 = 0
b) Now find numbers 𝒚and 𝒛(other than 𝒚 = 𝟎and 𝒛 = 𝟎) such that
𝑨 [
𝒚
𝒛
] = 𝟎
=A =b

Sol.
[
1 2
2 4
] [
𝑦
𝑧
] = 0
[
𝑦 + 2𝑧
2𝑦 + 4𝑧
] = 0
On comparing LHS and RHS
𝑦 = −2 𝑎𝑛𝑑 𝑧 = 1
c) If 𝑨 [
𝒚
𝒛
] = 𝟎, then explain why we must have
[
𝒚
𝒛
] = 𝜷 [
−𝟐
𝟏
]
for some number 𝜷.
Sol. From part (b) we get
[
𝒚
𝒛
] = [
−𝟐
𝟏
]
Hence for any number 𝛽, [
𝒚
𝒛
] = 𝜷 [
−𝟐
𝟏
] 𝑖𝑠 𝑡𝑟𝑢𝑒.
d) Take the numbers 𝒚and 𝒛from part (a). Without any calculation, explain why
𝑨 [
𝒚 + 𝟖
𝒛 − 𝟒
] = 𝒃
Sol.
𝐴 [
𝑦 + 8
𝑧 − 4
] = 𝑏
[1 2
2 4] [
𝑦 + 8
𝑧 − 4
] = [
1
2]
[1 2
2 4] [ 9
−4] = [
1
2]
Hence it is clear that LHS = RHS

Ques 3. a) Compute the 3 by 3 matrices 𝑴𝟏, 𝑴𝟐 𝑴𝟑where 𝑴𝟏 =
[
𝟏
𝟏
𝟏]
[𝟐 −𝟏 𝟑],
𝑴𝟐 =
[
𝟎
𝟏
−𝟑]
[𝟎 𝟐 −𝟏] and 𝑴𝟑 =
[
𝟎
𝟎
𝟏]
[𝟎 𝟎 𝟒].
𝑀1 = [
2 ∗ 1 1 ∗ (−1) 1 ∗ 3
1 ∗ 2 1 ∗ (−1) 1 ∗ 3
1 ∗ 2 1 ∗ (−1) 1 ∗ 3
]
𝑀1 = [
2 −1 3
2 −1 3
2 −1 3
]
𝑀2 = [
0 ∗ 0 0 ∗ 2 0 ∗ (−1)
1 ∗ 0 1 ∗ 2 1 ∗ (−1)
(−3) ∗ 0 (−3) ∗ 2 (−3) ∗ (−1)
]
𝑀2 = [
0 0 0
0 2 −1
0 −6 3
]
𝑀3 = [
0 ∗ 0 0 ∗ 0 0 ∗ 4
0 ∗ 0 0 ∗ 0 0 ∗ 4
1 ∗ 0 1 ∗ 0 1 ∗ 4
]
𝑀3 = [
0 0 0
0 0 0
0 0 4
]
b) Use Gaussian elimination and back substitution to solve 𝑨𝒙 = 𝒃, where
𝑨 = 𝑴𝟏 + 𝑴𝟐 + 𝑴𝟑and 𝒃 =
[
𝟏
𝟓
𝟏]
.
Sol.
𝐴 = 𝑀1 + 𝑀2 + 𝑀3
𝐴 = [
2 −1 3
2 −1 3
2 −1 3
] + [
0 0 0
0 2 −1
0 −6 3
] + [
0 0 0
0 0 0
0 0 4
]
𝐴 = [
2 −1 3
2 1 2
2 −7 10
]

𝑏 = [
1
5
1
]
𝐴𝑥 = 𝑏
[
2 −1 3
2 1 2
2 −7 10
] 𝑥 = [
1
5
1
]
𝑥 = 𝑏[𝐴]−1
[𝐴]−1 = 1
|𝐴| 𝐴𝑑𝑗[𝐴]
𝐴𝑑𝑗[𝐴] = (𝑀𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟𝑠[𝐴])𝑇
𝐴𝑑𝑗[𝐴] = [
24 −16 −16
−11 14 16
−5 2 4
]
𝑇
𝐴𝑑𝑗[𝐴] = [
24 −11 −5
−16 14 2
−16 16 4
]
|𝐴| = 2 ∗(24) + 2 ∗(11) + 2 ∗ (−5)
|𝐴| = 48 + 22 − 10
|𝐴| = 60
[𝐴]−1 = 1
60[
24 −11 −5
−16 14 2
−16 16 4
]
𝑥 = [
1
5
1
] . 1
60[
24 −11 −5
−16 14 2
−16 16 4
]
𝑥 = 1
12[
1
1
1
] [
24 −11 −5
−16 14 2
−16 16 4
]
𝑥 = 1
12[
24 −11 −5
−16 14 2
−16 16 4
]
𝑥 =
[
2 −11
12
−5
12
−4
3
7
6
1
6
−4
3
4
3
1
3 ]