Basic Linear Algebra Sample/Practice exam, questions and answers PDF

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Practice midterm exam for Basic Linear Algebra I (MATH 120) at MacEwan University. This practice exam is from the fall term with Dr. Strungaru. It includes detailed steps on how to answer each of the problems within the exam....

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Math 120 Midterm 1 Practice Problems Solutions

1.

Find all such values of the parameters b and c that the following linear system (i) will have exactly one solution; (ii) will have infinitely many solutions (iii) will be inconsistent: x 2y z 1

y x

3z c 3y bz

We write down the augmented matrix and perform row operations to convert to REF: 1 1 1 2 1 2 1 1 (𝑅 ) [0 1 −3 𝑐] 3 → 𝑅3 + 𝑅1 → [ 0 1 −3 𝑐 ] −1 −3 𝑏 2 0 −1 𝑏 + 1 3 1 1 1 2 (𝑅3 → 𝑅3 + 𝑅2 ) → [0 1 −3 𝑐 ] 0 0 𝑏−2 𝑐+3

We see that there are pivots in the first two columns. So 𝑥 and 𝑦 are leading variables for all values of 𝑏 and 𝑐. If 𝑏 − 2 ≠ 0, then we can divide the last row by 𝑏 − 2 to get 1 2 1 −1 [ 0 1 3 −𝑐 ] 𝑐+3 0 0 1 𝑏−2 In this case, all 3 variables are leading variables, so there is a unique solution if 𝑏 ≠ 2. If 𝑏 = 2, then the system can either be inconsistent or have infinitely many solutions, depending on the value of 𝑐 + 3. If 𝑏 = 2 and 𝑐 = −3, then the last row of the REF becomes [0 0 0 0] and so 𝑧 is a free variable and there are infinitely many solutions. If 𝑐 = 2 and 𝑐 ≠ −3, then we can divide the last row by 𝑐 + 3 to get [0 0 0 1]. This is equivalent to the equation 0 = 1, which is clearly false and so we obtain an inconsistent system in this case. 2.

(i)

Solve the following systems whose augmented matrix is given:  1 1 0 0  3 0 1 1 4 5     0 0 0 1 2  0 0 0 1 3   

0 0 −3 1 −1 0 0 −3 1 −1 5] 5 (𝑅 0 1 −1 4 0 1 −1 4 ] 4 → 𝑅4 − 𝑅3 ) → [ 0 0 [0 0 0 1 −2 0 1 −2 0 0 0 0 0 0 0 1 5 3 The augmented matrix is now in REF. The system is inconsistent because the last row states that 0 = 5. There are no solutions.

(ii)

é ê ê ê ê ë

1 2 0 0

1 0 1 2

0 0 0 0

0 0

1 0

3 3ù ú 1 4ú 0 3ú ú 0 0û

The augmented matrix is already in REF and it is equivalent to the system 𝑥1 + 2𝑥2 − 𝑥3 + 3𝑥5 = 3 𝑥3 + 2𝑥4 − 𝑥5 = 4 𝑥4 = 3 We see that 𝑥1 , 𝑥3 , and 𝑥4 are leading variables while 𝑥2 and 𝑥5 are free. Let 𝑥5 = 𝑡 and 𝑥2 = 𝑠 where 𝑠 and 𝑡 are arbitrary. Then we get 𝑥4 = 3 So, 𝑥3 = 4 − 2𝑥4 − 𝑥5 = 4 − 6 − 𝑡 = −2 − 𝑡 and 𝑥1 = 3 − 2𝑥2 + 𝑥3 − 3𝑥5 = 3 − 2𝑠 − (−2 − 𝑡) − 3𝑡 = 3 − 2𝑠 + 2 + 𝑡 − 3𝑡 = 5 − 2𝑠 − 2𝑡 The general solution is 𝑥1 = 5 − 2𝑠 − 2𝑡, 𝑥2 = 𝑠 , 𝑥3 = −2 − 𝑡, 𝑥4 = 3, 𝑥5 = 𝑡 where we may assign any values we like to 𝑠 and 𝑡. 3. a)

2

Do the vectors u = (1, -3) and v = (-1, 2) form a basis in R ? Explain. If they do form a basis, find the components of the vector a = (4, -1) in this basis.

A set of two vectors forms a basis for a plane if and only if they are not collinear. We can see by comparing the coefficients of the two vectors that neither one is a scalar multiple of the other. This means that they are not collinear which implies that they do form a basis for 𝑅 2 .

The components of the vector a = (4, -1) are the values of 𝑥1 and 𝑥2 that satisfy 𝒂 = 𝑥1 𝒖 + 𝑥2 𝒗. 4 −1 1 ]=[ ] So we solve the vector equation: 𝑥1 [ ] + 𝑥2 [ −1 2 −3 𝑥1 − 𝑥2 = 4 which is equivalent to the system −3𝑥1 + 2𝑥2 = −1 We can solve by converting the augmented matrix to REF: 4 1 −1 4 ] → [1 −1 4 ] (𝑅 → 𝑅 + 3𝑅 ) → [1 −1 ] [ 2 2 1 0 1 −11 0 −1 11 −3 2 −1 So 𝑥2 = −11 and 𝑥1 − 𝑥2 = 4 ⇒ 𝑥1 = 4 + 𝑥2 = 4 − 11 = −7 Thus, 𝒂 = −7𝒖 − 11𝒗 b) Do the vectors u = (1, 1, 0), v = (2, 1, -2), and w = (0, 1, 2) span

R3 ?

Explain.

A set of three vectors spans 𝑅 3 if and only if they are not coplanar. Although, we can see immediately that no two of these vectors are collinear (because no one of them is a multiple of any other), it is not obvious whether or not they are coplanar. They are coplanar if and only if they are linearly dependent so we check if the vector equation 𝑥1 𝒖 + 𝑥2 𝒗 + 𝑥3 𝒘 = 𝟎 has a nontrivial solution.

11 0 1 −1 0 ] is equivalent to the homogeneous system of equations 0 2 𝑥1 [ ] + 𝑥2 [ ] + 𝑥3 [ ] = [ 0 𝑥1 + 2𝑥2 =0 2 0 2 𝑥1 − 𝑥2 + 𝑥3 = 0 2𝑥2 + 2𝑥3 = 0 We solve by converting the coefficient matrix to REF: 1 2 0 1 2 0 1 2 0 1 [1 −1 1] (𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 ) → [ 0 −3 1] (𝑅2 ↔ 𝑅3 ) → [ 0 1 1] 2 0 2 2 0 −3 1 0 1 1 1 2 0 1 2 0 1 (𝑅3 → 𝑅3 + 3𝑅2 ) → [0 1 1] (𝑅3 → 4𝑅3 ) → [0 1 1] 0 0 1 0 0 4 Each variable is a pivot variable. There is no free variable. Thus, the homogeneous system has only the trivial solution, which implies that the three vectors are linearly independent.This implies that they form a basis for 𝑅 3. 4. a) Determine which of the following sets are bases in plane P shown in the diagram:  u, v ; a, b ;  a, v  u ;  b, u  a ;  a, b, v . Explain.

P

v

a

u b

A basis for a plane consists of two vectors in the plane that are not collinear. We can see immediately that u and v are not collinear, so {𝒖, 𝒗} is a basis. Also, a and b are not coplanar so {𝒂, 𝒃} is a basis. Using the triangle law, we can see that 𝒗 − 𝒖 = 𝒃, so {𝒂, 𝒗 − 𝒖} is a basis. Again, using the triangle law, we can see that 𝒖 − 𝒂 is collinear with 𝒃 (in fact, 𝒖 − 𝒂 = −14𝒃), so {𝒃, 𝒖 − 𝒂} is not a basis. {𝒂, 𝒃, 𝒗}is linearly dependent since it consists of more than 2 vectors in a plane. Thus, it is not a basis. b) Which of these sets are linearly independent? Substantiate you decisions.

The sets {𝒖, 𝒗}, {𝒂, 𝒃} and {𝒂, 𝒗 − 𝒖} are linearly dependent because they each consist of a pair of non-collinear vectors. {𝒃, 𝒖 − 𝒂} is not linearly independent because as already explained in (a), the two vectors are collinear .

{𝒂, 𝒃, 𝒗} is not linearly independent because any set of more than 2 vectors in a plane is linearly dependent. c) Find the coordinates of the vector p  12u +15v in each basis listed in question 4a).

12 The coordinates of 𝒑 = 12𝒖 + 15𝒗 in the basis {𝒖, 𝒗} are [ ] 15 1 2 Referring to the grid of vectors, we can see that 𝒖 = 𝒂 − 𝒃 and 𝒗 = 𝒂 + 𝒃, so 3 4 1 2 𝒑 = 12𝒖 + 15𝒗 = 12 (𝒂 − 𝒃) + 15 (𝒂 + 𝒃) = 12𝒂 − 3𝒃 + 15𝒂 + 10𝒃 = 27𝒂 − 7𝒃 3 4 27 The coordinates of 𝒑 in the basis {𝒂, 𝒃} are [ ] −7 As noted in (a), 𝒗 − 𝒖 = 𝒃, so {𝒂, 𝒗 − 𝒖} and {𝒂, 𝒃} are the same bases.

1   2  3     5. Show that the vectors a   0 , b   1 , c   1  form a basis for R3 and find the  2  0  1

11 coordinates of the vector 𝒖 = [ −2] with respect to this basis. 5

We solve the vector equation 𝑥1 𝒂 + 𝑥2 𝒃 + 𝑥3 𝒄 = 𝒖. If this equation has a unique solution, this will show that {𝒂, 𝒃, 𝒄} is a basis and the solution set will be the coordinates of 𝒖 in this basis. −2 3 11 1 𝑥1 [ 0] + 𝑥2 [ 1 ] + 𝑥3 [ 1 ] = [ −2] 5 2 0 −1 1 −2 [ 0 1 2 0

⇒ 𝑥1 − 2𝑥2 + 3𝑥3 = 11 𝑥2 + 𝑥3 = −2 2𝑥1 − 𝑥3 = 5 3 11 1 −2 3 11 1 −2 1 −2] → [ 0 1 1 −2 ] → [ 0 1 0 0 0 4 −7 −17 −1 5 1 −2 3 11 → [ 0 1 1 −2 ] 9 1 11 0 0

3 11 1 −2 ] −11 −9

So 𝑥3 = 11 , 𝑥2 + 𝑥3 = −2 ⇒ 𝑥2 = −2 − 11 = − 11 , 31 9 62 27 𝑥1 − 2𝑥2 + 3𝑥3 = 11 ⇒ 𝑥1 = 11 + 2 (− ) − 3 ( ) = 11 − − 11 11 11 11 121 − 62 − 27 32 = = 11 11 32 9 31 So, 𝒂 − 11 𝒃 + 11 𝒄 = 𝒖 9

11

9

31

Thus, the coordinates of 𝒖 in this basis are

32

[

11 31 − 11 9 11

]

6. For each of the following statements decide whether it is true (T) or false (F). Substantiate your decision. a. A set of three vectors cannot span R 2 . False. If the set includes two vectors that are not collinear, then it spans 𝑅 2. b. A system of 4 linear equations in 3 variables can have a unique solution. True. The coefficient matrix and the augmented matrix could both have rank 3. In this case the system would be consistent and since each variable would be a pivot variable, there would be a unique solution. c. A system of 2 linear equations in 3 variables can have a unique solution. False. If the system is consistent, then it must have infinitely many solutions because there must be at least one free variable. d. A system of 3 linear equations in 3 variables must have a unique solution. False. It will have unique solution only when the coefficient matrix has rank 3. If the rank of the coefficient matrix is less than 3 and the system is consistent, then it will have infinitely many solutions. It is also possible to have an inconsistent system of 3 linear equations in 3 variables.

7. Select (encircle) the right answer for each question. Do not explain your choice. (i)

(ii)

If for some system in 5 variables the rank of the coefficient matrix is equal to 4, then (a) The system is homogeneous (b) The system is inconsistent (c) The system does not have a unique solution (d) None of the above (a,b,c) is true. A homogeneous system in 4 variables with the rank of the coefficient matrix equal to 4 (a) Will have only the trivial solution (b) Will be inconsistent (c) Will have a nontrivial solution (d) Will have 4 solutions....