Linear Algebra Worksheet

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avygba1

Mathematics

Description

Do not skip any steps in your calculations or arguments. Justify your answers rigorously

the written exercises part must be a single PDF file.

show calcutalions step by step, us mathlab and attach a screenshot on the pdf.Make the answers as clear as possible.


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MATH 2318 SPRING 2022 HOMEWORK 01 Do not skip any steps in your calculations or arguments. Justify your answers rigorously! Written exercises. 1. (12 points) Find the point of intersection of the lines x1 + 2x2 = −13 and 3x1 − 2x2 = 1. Formulate the problem as a linear system, then use the row reduction process. 2. (10 points) Let  1  0   0 0 3 1 0 0  0 −2 −7 0 3 6   2  1 0 0 1 −2 be the augmented matrix of a linear system. Complete the row reduction process and describe the solution set of the linear system. 3. (10 points) Solve the linear system   2x1  3x1 − 6x3 = −8 x2 + 2x3 = 3 + 6x2 − 2x3 = −4 4. (8 points total, 2 points each) Answer “True” or “False” to each the following. Justify your answers. (a) Two matrices are called row equivalent if they have the same number of rows. (b) Elementary row operations on an augmented matrix never change the solution set of the corresponding linear system. (c) Two equivalent linear systems can have different solution sets. (d) A consistent system of linear equations has one or more solutions. 5. (8 points total, 2 points each) Determine which matrices are in reduced echelon form and which others are in echelon form only.         0 1 1 1 1 0 0 0 0 1 0 1 1 1 0 0 0   1 2 0 0  , (d) 0 0 1 1 1 . (a) 0 1 1 1 , (b) 0 2 0 0 , (c)  0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 Distribution of this material without instructor’s written permission is prohibited. 1 6. (12 points) Find the general solution of a linear system with the augmented matrix   1 −2 −1 4 . −2 4 −5 6 7. (10 points total, 2 points each) Answer “True” or “False” to each the following. Justify your answers. (a) Reduced echelon form of a matrix is unique. (b) If every column of an augmented matrix contains a pivot, then the corresponding linear system is consistent. (c) The pivot positions in a matrix depend on whether row interchanges are used in the row reduction process. (d) A general solution is an explicit description of all solutions of a linear system. (e) Whenever a system has free variables, the solution set contains infinitely many solutions. 8. (10 points) Suppose a 3 × 5 coefficient matrix of a linear system has three pivot columns. Is the system consistent? Justify your answer. 9. (10 points) A system of linear equations with fewer equations than unknowns is called an “underdetermined” system. Can such a system have a unique solution? Justify your answer. Matlab exercise. 1. (10 points) Let A be a k × 4 matrix whose entries are random numbers between 0 and 1. Write a script that prints on the screen the reduced echelon form of A for k = 2, 3, 4, 5. Distribution of this material without instructor’s written permission is prohibited. 2
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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.talk a look at these 3 files. IF you want me to handwrite the word doc OR simply save it as a PDF, I can do that. Handwritten won't be as neat.

Question #1
1
[
3

2 −13
|
]
−2
1

𝑔𝑖𝑣𝑒𝑛

1
[
0

2 −13
|
]
−8
40

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟1𝑐1. 𝑛𝑒𝑤 𝑟𝑜𝑤 2 = −3 ∗ 𝑟𝑜𝑤1 + 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑟𝑜𝑤 2 (𝑟2′ = −3𝑟1 + 𝑟2)

1
[
0

2 −13
|
]
1
5

𝑛𝑒𝑤 𝑟𝑜𝑤 2 = 𝑜𝑙𝑑 𝑟𝑜𝑤 2⁄−8

1 0 −3
[
|
]
0 1 5
𝑥1
−3
[𝑥 ] = [ ]
5
2

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟2𝑐2. 𝑛𝑒𝑤 𝑟𝑜𝑤 1 = −2 ∗ 𝑟𝑜𝑤2 + 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑟𝑜𝑤 1 (𝑟1′ = −2𝑟2 + 𝑟1)

Question #2
1
0
[
0
0

3
1
0
0

0
0
1
0

−2 −7
3 6
| ]
0 2
1 −2

𝑔𝑖𝑣𝑒𝑛

1
0
[
0
0

0
1
0
0

0
0
1
0

−11 −25
3
6
|
]
0
2
1
−2

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟2𝑐2. 𝑛𝑒𝑤 𝑟𝑜𝑤 1 = −3 ∗ 𝑟𝑜𝑤2 + 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑟𝑜𝑤1

1
0
[
0
0

0
1
0
0

0
0
1
0

0 −47
0 12
|
]
0 2
1 −2

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟4𝑐4. 𝑟1′ = 11𝑟4 + 𝑟1

−47
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 #1
12
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 #2
[
]=[
]
2
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 #3
−2
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 #4

Question #3
2 0 −6 −8
[0 1 2 | 3 ]
3 6 −2 −4

𝑔𝑖𝑣𝑒𝑛

1 0 −3 4
[0 1 2 | 3 ]
3 6 −2 −4

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟1𝑐1. 𝑟1′ = 𝑟1 ⁄ 2

1 0 −3 −4
[0 1 2 | 3 ]
0 6 7 8

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟1𝑐1. 𝑟3′ = −3𝑟1 + 𝑟3

1 0 −3 −4
[0 1 2 | 3 ]
0 0 −5 −10

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟2𝑐2. 𝑟3′ = −6𝑟2 + 𝑟3

𝑟2′ = −3𝑟4 + 𝑟2

1 0 −3 −4
[0 1 2 | 3 ]
0 0 1 2
1 0 0 2
[0 1 0 | −1]
0 0 1 2
𝑥1
2
𝑥
[ 2 ] = [−1]
𝑥3
2

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟3𝑐3. 𝑟3′ = 𝑟3 ⁄−5

𝑝𝑖𝑣𝑜𝑡 𝑜𝑛 𝑟3𝑐3. 𝑟1′ = 3𝑟3 + 𝑟1

𝑟2′ = −2𝑟3 + 𝑟2

Question #4
a) FALSE. By definition, two matrices are row equivalent if elementary row operations can be
performed on one matrix to obtain the other. Two matrices with the same number of rows
might not be row equivalent. Example
1
[
3

2
1
] 𝑎𝑛𝑑 [
4
4

2
]
4

Are not row equivalent even though they both have 2 rows because there are no row operations
to obtain the 2nd from the 1st matrix.
b) TRUE. Elementary row operations never change the solution set because those operations are
equivalent to algebra being performed on both sides of an equation.
c) FALSE. By definition, two linear systems are equivalent ONLY if they have the same solution set.
d) TRUE. By definition, an inconsistent system has zero solutions. A consistent set has 1 or more
solutions. An independent consistent system has exactly 1 solution.

Question #5
Echelon form means
1) The leading coefficient in each row is to the left of the leading coefficient of the row
below it.
2) Rows with all zeros are at the bottom
Reduced row Echelon form means echelon AND
3) The leading coefficient in each row is the only nonzero in that column
Examples
1 4 −3
[0 1 2 ] 𝑛𝑜𝑡 𝑒𝑐ℎ𝑒𝑙𝑜𝑛. 𝑁𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑎 (1) 𝑛𝑜𝑟 (2) 𝑎𝑟𝑒 𝑚𝑒𝑡
3 6 −2
1 4 −3
[0 2 2 ] 𝑒𝑐ℎ𝑒𝑙𝑜�...


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