Linear Algebra Assignment 1 2021, The questions and answers PDF

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 1 Which of the following is a linear equation in x, y, and z?

1. 2. 3. 4.

x − e2 y = 3z 1

2π ln (e z ) − 2y + z = ln(3) − x √y −2 + 4y − 2z = 7x y + 4y − 2z = 7x 3

Solu Soluti ti tion on Linear Equations:

Properties of natural logs:



Highest power of variables is one.



No products or roots of variables.



Variables do not appear as arguments of trig, logarithmic or exponential functions. .

1. ln(x y ) = yln x 2. ln(ex ) = x 3. ln(e) = 1 4. eln(x) = x

Exponent(s) laws: 1. a−m =

1 1 = a−m or am am

Note: I only chose properties of natural logs and exponents laws relevant to the concepts in question.

SOL SOLU UTION

1. 2.

x − e2 y = 3z ⇒ x − 7.38 … y = 3z 1

2π ln (e z ) − 2y + z = ln(3) − x ⇒ 2πz −1 − 2y + z = ln(3) 1

Note: ln (e z ) = 3.

4.

√y −2

1 1 1 = z −1 or ln(e) = = z −1 z z z

+ 4y − 2z = 7x

y + 4y − 2z =

∵ ln(e) = 1and ln(3) = 1.098 ..

7x 3

QU QUESTI ESTI ESTION ON 2 © 2021 JTS Maths Tutoring Keeping you mathematically informed

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Which of the following is a linear equation in x, y, and z? 1. 2. 3. 4.

1.

2. 3.

4.

x 2 − e2 y = 3z

2π ln(ez ) − 2y + z = ln(3) − x √y −1 + 4y − 2z = 7x y + 4y − 2z = 7x3

SOL SOLUTION UTION

x2

−e2 y = 3z

√y −1

+ 4y − 2z = 7x

2π ln(ez ) − 2y + z = ln(3) − x ⇒ 2πz − 2y + y = ln(3) − x y + 4y − 2z =

7x3

QU QUESTI ESTI ESTION ON 3 Which of the following is a nonlinear equation in x, y, and z? 1. 2. 3. 4.

1. 2. 3. 4.

y + 4y − 2z = 7x

2π ln(e−z ) − 2y + z = ln(3) − x ⇒ −2πz − 2y + z = ln(3) − x 7x − 4y − 2z = 0 x − exy = 3z

y + 4y − 2z = 7x

SOL SOLUTION UTION

2π ln(e−z ) − 2y + z = ln(3) − x ⇒ −2πz − 2y + z = ln(3) − x 7x − 4y − 2z = 0 exy x−

exy

= 3z

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MAT1503 Ass 1 2021 Memorandum

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MAT1503 Ass 1 2021 Memorandum

In the following three questions, draw a table of logical operations in order to boil down statements into digestible operations through the corresponding logical formulas. QU QUESTI ESTI ESTION ON 4 Any homogeneous linear system has one solution or no solution.

1. 2. 3. 4. 5.

T ∧ F, where T stands for True and F stands for False

(F ∨ F ) ∧ ⌝ T, where ⌝ is the symbol for logical negation F ∧ (⌝ F ∨ T) (T ∨ F ) ∧ F (T ∨ F ) ∨ F SOL SOLUTION UTION Log Logical ical C Conne onne onnectives ctives

dec declarativ larativ larative e sta statemen temen tements ts tha thatt are eithe eitherr true or fa false lse but ca can’t n’t be bot both h).

Logical connectives are operators that are used to join one or more prepositions (propositions ar are e

Con Connecti necti nective ve

Con Connecti necti nective ve W Word ord

Sym Symbol bol

Con Conjunct junct junction ion

And

Disj Disjunct unct unction ion

Or

Con Condition dition ditional al

If--then

∨

(im (implicat plicat plication) ion) Bico Biconditi nditi nditional onal

If and only if (iff)

Neg Negation ation

Not It is not the case that

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∧

→

↔

⌝ or ~ or ‘ or -

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Con Connecti necti nective ve ∧ (an (and) d)

∨ (or)

Des Descript cript cription ion If p and q are two propositions, then the conjunction of p and q or p ∧ q will translate to:  True iff both p and q are true.  Fal False se when both p and q are false and when either one of p or q is true or false. If p and q are two propositions, then the disjunction of p and q or p ∨ q will translate to:  True iff one of p or q or both are true.  Fal False se when both p and q are false.

If p is a propositional, then the Neg Negation ation (⌝ or ~ o orr ‘or -) negation of p will translate to:  True when p is false  Fal False se when p is true.

MAT1503 Ass 1 2021 Memorandum

Tru Truth th Tab Table le INP INPUTS UTS p

q

T T F F

T F T F

INP INPUTS UTS p

q

T T F F

T F T F

p T F

⌝p F T

OU OUTPU TPU TPUT T p ∧q T F F F

OU OUTPU TPU TPUT T p∨q T T T F

No Note: te: I on only ly dea dealt lt wi with th the ab above ove llogical ogical conn connective ective ectivess rele relevant vant to the conc concepts epts in que question. stion. Now we evaluate our given statements or propositions based on the truth tables above. Rea Reality: lity: Any homogeneous linear system has one or infinitely many solutions. Give Given n sta statemen temen tementt (pr (propos opos oposition) ition) ition):: Any homogeneous linear system has one solution or no solution. The given statement is a disjunction statement which will translate to True (T) ∵ one of the given conditions is True (T). So, we will evaluate all the given options, the one with True (T) as the final output is the correct option. T∧F =F

Opt Option ion 1

Opt Option ion 2 (F ∨ F) ∧ ¬T =F ∧ F =F

Opt Option ion 4 (T ∨ F) ∧ F =T∧F =F

Opt Option ion 5 (T ∨ F) ∨ F =T∨F =T

Opt Option ion 3 F ∧ (¬F ∨ T) = F ∧ (T ∨ T) =F∧T =F

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 5 Any homogeneous linear system has no solution or infinitely many solutions.

1. 2. 3. 4. 5.

T ∧ F, where T stands for True and F stands for False

(F ∨ F ) ∧ ⌝ T, where ⌝ is the symbol for logical negation F ∧ (⌝ F ∨ T)

(⌝ T ∨ F ) ∨ (⌝ F) (T ∧⌝ F ) ∧ (⌝ F)

SOL SOLUTION UTION The given statement is a disjunction statement which will translate to True (T) ∵ one of the given conditions is True (T). So, we will evaluate all the given options, the one with True (T) as the final output is the correct option.

Opt Option ion 1

Opt Option ion 2 (F ∨ F) ⋀ ⌝ T =F ∧ F =F

T∧F =F

Opt Option ion 4 (⌝ (⌝ T ∨ F) ∨ F) =F∨F∨T =F∨T =T

Opt Option ion 5 (T ∧⌝ T ) ∧ (⌝ F) =T∧F∧T =F∧T =F

Opt Option ion 3 F ∧ (⌝ F ∨ T) = F ∧ (T ∨ T) =F∧T =F

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 6 Any homogeneous linear system has one solution or infinitely many solutions.

1. 2. 3. 4. 5.

T ∧ F, where T stands for True and F stands for False (⌝ T ∧⌝ F ) ∨ (⌝ F) F ∧ (⌝ F ∨ T) (T ∨ F ) ∧ F

(F ∨ F ) ∧⌝ T, where ⌝ is the symbol for logical negation SOL SOLUTION UTION

The given statement is a disjunction statement which will translate to True (T) ∵ one of the given conditions is True (T). So, we will evaluate all the given options, the one with True (T) as the final output is the correct option. T∧F =F

Opt Option ion 1

Opt Option ion 4 (T ∨ F ) ∧ F =F∧F =F

Opt Option ion 2 (⌝ T ∧⌝ F ) ∨ (⌝ F) =F ∧ T ∨ T =F ∨ T =T

Opt Option ion 3 F ∧ (⌝ F ∨ T) = F ∧ (T ∨ T) =F∧T =F

Opt Option ion 5 (F ∨ F ) ∧⌝ T =F∧F =F

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 7 Determine which of the following is the solution set of the linear equation below: 3x − y + z = 2

1. 2.

3. 4.

1 {(x, y, z): x = (t − s + 2), y = t, z = s with s, t ∈ ℝ} 3

1 {(x, y, z): x = (t + s + 2), y = t, z = s with s, t ∈ ℝ} 3 1 {(x, y, z): x = (t − s − 2), y = t, z = s with s, t ∈ ℝ} 3

1 {(x, y, z): x = − (2 + t −), y = t, z = s with s, t ∈ ℝ} 3 SOL SOLUTION UTION

To find the solution set of one linear system of equation involving two or more variables, solve for one of the variables (the leading variable) in terms of the other variables. No Note: te:   

The leading variable is a dep depende ende endent nt var variable iable since its value depends on the values of the other variables. The other variables are free (or indepe independe nde ndent) nt) vvariable ariable ariabless since they can take on any real value. Free variables can be assigned arbitrary values (parameters) - parameters can be any letters of the alphabet which do not conflict with the names of the given unknowns. 3x − y + z = 2

3x = y − z + 2

[solving x in terms of y and z]

1 x = (y − z + 2) [divide both sides by 3] 3 1 x = (t − s + 2) [assign parameters to free 3 variables y = t and z = s, s, t are any real numbers]

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 8 Consider the system obtained from the augmented matrix: [ Choose the correct statement(s):

1. The system has no solution if ae = bd

a b c | ] d ef

2. 𝑇he system has exactly one solution whenever ae ≠ bd 3. The system is inconsistent for af ≠ ab

4. The system has infinitely many solutions if a = c b f

5. The system has no solution if ae = bf or impossible SOL SOLUTION UTION

Num Number ber o off Solu Solution tion tionss of a Nonh Nonhomoge omoge omogeneou neou neouss Sys System tem of Equa Equations tions Using The Dete Determina rmina rminant nt

Suppose we have two equations and two unknowns: ax + by = c dx + ey = f

The above are two equations represent two lines in the xy-plane. Solving these simultaneous equations yields three (3) solutions: one solution, no solution, and infinite solutions. Graphically representing the above we have: y

y

y

x

x

x

No solut solution ion

Infi Infinite nite ssolution olution olutionss

One solut solution ion

In xyz-plane, we also have one solution (planes intersecting at a point), no solution (no common intersection) and infinitely many solutions (planes intersecting at a line or intersecting at a plane). © 2021 JTS Maths Tutoring Keeping you mathematically informed

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MAT1503 Ass 1 2021 Memorandum

If we rewrite the lines using the slope-point formula, we will have: a c y = − x + and b b

d f y= − x+ e e

a d and − as slopes b e (or gradients)respectively. The two lines have −

Graphically, it can be noted that the equations (lines) have: 





𝐚 𝐮𝐧𝐢𝐪𝐮𝐞 (𝐨𝐧𝐞)𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 when they intersect at a point. This occurs when the a d slopes (gradients) of the lines are not equal, i. e. , ≠ . b e 𝐧𝐨 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 when the lines are parallel. This occurs when the slopes (gradients) a d are equal, i. e. , = . b e 𝐢𝐧𝐟𝐢𝐧𝐢𝐭𝐞𝐥𝐲 𝐦𝐚𝐧𝐲 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧𝐬 when the lines are identical. This occurs when the a d slopes (gradients) are equal, i. e. , = . b e

Let the coefficient matrix be A = [

a b ] , det(𝐴) = ae − db or ae − bd. Assume d e

a d = , i. e. , the slopes are equal. From the notes above, the b e lines are either parallel (𝐭𝐡𝐞𝐫𝐞 𝐢𝐬 𝐧𝐨 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧) or are identical (𝐭𝐡𝐞𝐫𝐞 𝐢𝐬 𝐢𝐧𝐟𝐢𝐧𝐢𝐭𝐞𝐥𝐲 𝐦𝐚𝐧𝐲 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧𝐬). If ae − bd ≠ 0, then the slopes are not equal, i. e. , the lines intersect at exactly one point. This boils down to the following conclusion:

ae − bd = 0 ⇒ ae = bd ⇒

Using the dete determin rmin rminant ant ant, an n x n nonhomogeneous system has the following solutions: 

Uni Unique que ssolution olution (exa (exactly ctly one ssolution olution olution))



No solut solution ion



Infi Infinitely nitely man many y solu solutions tions

Iff its determinant is nonzero, i. e. , ae − db ≠ 0 ⇒ ae ≠ db ⇒

If the determinant is zero, i. e. , ae − db = 0 ⇒ ae = db ⇒

If the determinant is zero, i. e. , ae − db = 0 ⇒ ae = db ⇒

a d c ≠ = b e f

a d c = ≠ b e f

a d c = = b e f

No Note te te: There are two (2) correct options on this Question, hence the instruction: Choose the correct statement(s) not Choose the correct statement.

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QU QUESTI ESTI ESTION ON 9 Consider the system obtained from the augmented matrix below: [

Choose the correct option:

1.

a bc | ] d ef

The system has infinitely many solutions if

a b e = = c d f

2. The system has exactly one solution whenever af ≠ bd 3. The system is inconsistent for af ≠ ab 4. The system has no solution if ae = bd 5. The system has no solution if ae = bf

SOL SOLUTION UTION

Use notes on Question 8 to answer this Question.

QU QUESTI ESTI ESTION ON 10 {

Consider the following linear system:

2x − 3y = −1 2x − 3y = 1

1. x = 0 and y = 0 staisfy the system.

2. The system has exactly one solution. 3. The system is inconsistent .

4. The system has infinitely many solution.

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MAT1503 Ass 1 2021 Memorandum

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MAT1503 Ass 1 2021 Memorandum

SOL SOLUTION UTION

By eelimina limina limination tion

2x − 3y = −1

−(2x − 3y = 1)

0 = −2

1 2

0 = −2 is a contradictory statement, hence the system has no solution.

Usin Using g the deter determina mina minant nt

2x − 3y = −1

Augmented matrix: [

2x − 3y = 1

2 −3 2 −3 ]⇒| | = −6 + 6 = 0 2 −3 2 −3

By substi substitution tution

From

2

2(

2

2x − 3y = 1

x=

Substitute x =

1

2x − 3y = −1 1 + 3y 2

1 + 3y into 2

1 + 3y ) − 3y = −1 2

1

1 + 3y − 3y = −1

0 = −2

An n x n homogeneous systesm has no solution iff its determinant is zero.

0 = −2 is a contradictory statement, hence, the system has no solution.

Gau Gauss ss elim eliminat inat ination ion

Gra Graphica phica phicallly

2x − 3y = −1 and 2x − 3y = 1

y

Write the above system of equations as an augmented matrix and use Gauss elimination: 2 −3 −1 ] 1 [ R → R1 | 2 −3 1 2 1

3 1 [1 − 2|− 2] 2R1 − R 2 → R2 2 −3 1

2x − 3y = −1 2x − 3y = 1

3 1 [1 − 2| − 2 ] 0 0 −2

The equation that corresponds to the last row of the augmented matrix is 0x + 0y = −2

x

The lines are parallel (but not identical); hence the system has no solution.

Since this equation is not satisfied by any values of x and y, the system has no solution.

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QU QUESTI ESTI ESTION ON 11 x − y + 3z = 1 2x − y + z = −1 Consider the following linear system: { x − 3y − z = 2

Which of the following is the augmented matrix for the above given system? 1.

2.

3.

4.

5.

1 −1 [2 −1 1 3

3 1 1 | −1 ] −1 2

1 −1 [2 −1 1 3

3 1 1 | −1 ] −1 3

1 −1 [2 −1 1 3 1 −1 [2 −1 1 3

3 1 1 | −1 ] −1 1

3 1 1 | −1 ] −1 0

1 −1 3 1 [2 −1 1| −1 ] 1 −3 −1 2

SOL SOLUTION UTION

An aug augmente mente mented d ma matrix trix is a matrix made up of the coefficients of the variables (or unknowns) and the column of the results (or answers). The coefficients of the variables and the answers are separated by a line called an augment. Let us suppose we have the following system:

ax1 − bx1 + cx1 = a dx2 − ex2 + fx2 = −b gx3 + hx3 − ix3 = c

then, the augmented matrix of the above system is: a −b [d −e g h

c a f | −b ] −i c

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MAT1503 Ass 1 2021 Memorandum

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MAT1503 Ass 1 2021 Memorandum

QU QUESTI ESTI ESTION ON 12 Assume that we are given a coefficient matrix A of the linear equations AX = b, where 1 b = [2

],X = [

x12

0 2 1 5 2 6] ] and A = [ 1 0 0

x3 4 Solve for the variable x2 and x3 in the above system. 1. The system has infiitely many solutions 2. The system has exactly one solution 3. 𝑇he system is inconsistent

4. x2 = 0 and x3 = 1 satisfy the system 5. x2 = 1 = x3

SOL SOLUTION UTION

Usin Using g the m matrix atrix eq equation uation 𝐀𝐗 = 𝐛 0 1 2 1 2 | = −4 | 2 5 6| = 1 | 5 6 1 0 0

A𝐗 = b ⇒ 𝐗 = A b

1 𝐗 = A−𝟏 b = adj(A)b |A| 1 𝐗=− [ 4 𝐗=

[

−

0

6 4 5 4

0 6 −5

−𝟏

0 −4 1 −2 4] [2] [ 1 −2 4

0

2 4 1 − 4

4 4 4 1 − [ 2] 4 2 4 4]

𝐗=

5 0 1 5

6 0 2 6

2 1 0 2

5 0 1 5]

[

0 3 − 2 5 4

0 1 1 −1 1 [2] 2 1 1 4 − 4 2]

0+0+4 4 3 9 − +1−4 − = 𝐗= 2 2 5 1 11 [ 4 − 2 + 2] [ 4 ]

∴ x1 = 4, x2 = −

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9 11 and x3 = 4 2

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Usin Using g sub substitu stitu stitution tion

Using Gauss-Jordan 02 15 26 2 1 ] R1 ↔ R 3 [ |4 1 0 0 1 0 0 4 [ 2 5 6 2] 2R1 − R2 → R 2 | 1 0 1 2

From the given matrix equation, we obtain the following linear system of equations:

0

1

0

[0

0

0

1 1

1

[0

0

0

1

1

[0

[

1 0 0

0

0

0 1 0

x2 + 2x3 = 1

x1

1

2

2x1 + 5x2 + 6x2 = 2

1 0 0 4 1 [ 0 −5 −6 |6] − R 2 → R 2 5 0 1 21 1

MAT1503 Ass 1 2021 Memorandum

=4

3

It can be noted that x1 = 4. We can back

0 4 6| 6 ] R − R → R 3 3 5−5 2 2 1

substitute into the second equation to get: x2 + 2x3 = 1

8 + 5x2 + 6x3 = 2

0 4 6 −6 5 5 | 5 − R3 → R3 4 11 4 −5 − 5 ]

Now we have:

0 4 6 6 −6 − R 3 + R 2 → R2 5 | 5 5 1 11 ] 4

4 0 9 9 11 0|− 2 ∴ x1 = 4, x2 = − and x3 = 4 2 1 11 4]

x1

x2 + 2x3 = 1

=4

5x2 + 6x2 = −6

1

4

From the first equation: x2 = 1 − 2x3

Substitute x2 = 1 − 2x3 into the fourth equation, we obtain:

5(1 − 2x3 ) + 6x3 = −6

5 − 10x3 + 6x3 = −6

−4x3 = −11