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Quiz 6: Solving Linear Systems

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Question 1

 
 
Select all the matrices which are in row echelon form.
a) ⌊ ⌋ 3 1 1 - 1 2 ⌈0 1 0 0 1⌉ 0 0 0 - 2 7   b) ⌊ ⌋ 1 1 1 - 1 ⌈1 1 0 0 ⌉ 0 0 0 0
c) ⌊ ⌋ 1 1 1 4 ⌈0 0 0 0⌉ 0 1 0 0   d) [ ] 0 0 0 0
e) ⌊ ⌋ 0 1 0 0 ⌈0 0 1 0⌉ 0 0 0 1

 

There is at least one mistake.
For example, choice (a) should be false.
This is not in row echelon form because the leading terms in rows 1 and 3 are not equal to 1.
There is at least one mistake.
For example, choice (b) should be false.
This is not in row echelon form because the leading term in the second row is not further to the right than the leading term in the first row.
There is at least one mistake.
For example, choice (c) should be false.
This is not in row echelon form because the row of zeros is not underneath all the non-zero rows.
There is at least one mistake.
For example, choice (d) should be true.
There is nothing in the definition of row echelon form that says the matrix must have some non-zero entries. The zero matrix of any size is always in row echelon form.
There is at least one mistake.
For example, choice (e) should be true.
Your answers are correct
  1. False. This is not in row echelon form because the leading terms in rows 1 and 3 are not equal to 1.
  2. False. This is not in row echelon form because the leading term in the second row is not further to the right than the leading term in the first row.
  3. False. This is not in row echelon form because the row of zeros is not underneath all the non-zero rows.
  4. True. There is nothing in the definition of row echelon form that says the matrix must have some non-zero entries. The zero matrix of any size is always in row echelon form.
  5. True.
 

Question 2

 
 
Select all the matrices which are in reduced row echelon form.
a) ⌊ ⌋ 1 0 1 0 ||0 1 0 2|| ⌈0 0 1 7⌉ 0 0 0 0   b) ⌊ ⌋ ⌈1 0 0 0⌉ 0 0 0 0 0 1 0 0
c) ⌊0 0 0 0 0 0⌋ ⌈0 0 0 0 0 0⌉ 0 0 0 0 0 0   d) [ ] 0 1 0 2 0 0 1 11

 

There is at least one mistake.
For example, choice (a) should be false.
The leading 1 in row 3 has a non-zero entry above it in the same column, so this matrix is not in reduced row echelon form.
There is at least one mistake.
For example, choice (b) should be false.
This is not in reduced row echelon form because the row of zeros is not underneath all the non-zero rows.
There is at least one mistake.
For example, choice (c) should be true.
There is at least one mistake.
For example, choice (d) should be true.
Your answers are correct
  1. False. The leading 1 in row 3 has a non-zero entry above it in the same column, so this matrix is not in reduced row echelon form.
  2. False. This is not in reduced row echelon form because the row of zeros is not underneath all the non-zero rows.
  3. True.
  4. True.
 

Question 3

 
 
One more row operation must be applied to ⌊ ⌋ 1 1 4 5 ⌈ 0 0 1 4⌉ 0 1 1 0 in order to obtain a matrix in row echelon form. Which operation is it?
a) An operation on row 2: R2 - R3   b) An operation on row 3: R3 - R1
c) An operation on row 3: R3 - R2   d) Swap rows 2 and 3: R2 ↔ R3.

 

Not correct. Choice (a) is false.
Do the row operation and you’ll see why the resulting matrix is not in row echelon form!
Not correct. Choice (b) is false.
Do the row operation and you’ll see why the resulting matrix is not in row echelon form!
Not correct. Choice (c) is false.
Do the row operation and you’ll see why the resulting matrix is not in row echelon form!
Your answer is correct.
 

Question 4

 
 
In trying to find out if the lines (in 3D space)
⌊1⌋ ⌊2⌋ ⌊2⌋ ⌊1⌋ x = ⌈3⌉ + s⌈1⌉ and x = ⌈5⌉ + t⌈1⌉ 4 2 1 3
have a common point of intersection, we solve a system of 3 equations in the two unknowns s and t. After applying a number of elementary row operations (starting with the augmented matrix) we obtain ⌊ | ⌋ 1 0|- 1 ⌈ 0 1|- 3 ⌉ 0 0 1. What conclusion can be drawn about the point of intersection?
a) The point of intersection is (-1,-3,1).   b) The point of intersection is (1,3,-1).
c) The point of intersection is (-1,-3,0).   d) The point of intersection is (1,3,0).
e) The two lines do not intersect.   f) There is not enough information given to answer the question.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Your answer is correct.
The third row of the matrix corresponds to the equation 0s + 0t = 1, that is, that 0 = 1. This shows that the system is inconsistent. There are no solutions for s and t and hence the two lines do not meet.
Not correct. Choice (f) is false.
 

Question 5

 
 
Consider the matrix ⌊1 0 0⌋ ⌈0 2 1⌉ 3 1 1. Which is the most sensible elementary row operation to perform next, if we want to progress towards a matrix in row echelon form?
a) An operation on row 3: R3 - 3R1.   b) An operation on row 3: R3 -1 2R2.
c) Swap rows 2 and 3.   d) An operation on row 2: R2 - 2R3.

 

Your answer is correct.
In working towards row echelon form, we move through the matrix from left to right and from top to bottom. Once the leading entry is in place in row 1, we make all entries underneath it equal to zero. Therefore we should clear the entry in row 3, column 1 before we do anything else.
Not correct. Choice (b) is false.
This will make a zero in the second entry of row 3, but this is not necessary for row echelon form.
Not correct. Choice (c) is false.
This will not give us an efficient path to row echelon form because the position currently occupied by 3 in the first column will still need to be cleared to zero.
Not correct. Choice (d) is false.
 

Question 6

 
 
Consider the matrix ⌊ ⌋ 1 0 0 1 ⌈0 0 3 1⌉ 0 1 1 0. Which is the most sensible elementary row operation to perform next, if we want to progress towards a matrix in reduced row echelon form?
a) An operation on row 2: 13R2.   b) An operation on row 2: R2 - 3R3.
c) An operation on row 1: R1 - R2.   d) Swap rows 2 and 3.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
This will produce the matrix ⌊ ⌋ 1 0 0 1 ⌈0 1 1 0⌉ 0 0 3 1. Reduced row echelon form can be obtained with just two more elementary row operation. (Can you see what they are?)
 

Question 7

 
 
For which value of k is the system
x + 2y = k
2x - ky = 4
inconsistent (that is, has no solution)?
a) k = 0   b) k = -4
c) k = 2   d) k = -1

 

Not correct. Choice (a) is false.
Hint: write down the augmented matrix of the system and apply a row operation to work towards row echelon form. Then search for the value of k that makes the system inconsistent.
Your answer is correct.
The augmented matrix is row equivalent to the matrix [ 1 2 | k ] 0 - k - 4|4- 2k. Therefore if k = -4, the last row becomes [0 0 | 12], indicating an inconsistent system.
Not correct. Choice (c) is false.
Hint: write down the augmented matrix of the system and apply a row operation to work towards row echelon form. Then search for the value of k that makes the system inconsistent.
Not correct. Choice (d) is false.
Hint: write down the augmented matrix of the system and apply a row operation to work towards row echelon form. Then search for the value of k that makes the system inconsistent.
 

Question 8

 
 
For which value of a does the system
x + ay + z = 1
ax + y + 2z = 0
x - 2y + z = 1
have infinitely many solutions?
a) a = 1   b) a = -3
c) a = 5   d) a = -2

 

Not correct. Choice (a) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations (working towards row echelon form) until the matrix is simplified. (The matrix will continue to contain various terms involving a.) Then search for the value of a that gives just two non-zero rows in the matrix.
Not correct. Choice (b) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations (working towards row echelon form) until the matrix is simplified. (The matrix will continue to contain various terms involving a.) Then search for the value of a that gives just two non-zero rows in the matrix.
Not correct. Choice (c) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations (working towards row echelon form) until the matrix is simplified. (The matrix will continue to contain various terms involving a.) Then search for the value of a that gives just two non-zero rows in the matrix.
Your answer is correct.
The augmented matrix of the system can be reduced to
⌊ 1 a 1 | 1 ⌋ ⌈ 0 1- a2 2 - a|- a ⌉. 0 - 2- a 0 | 0
Therefore when a = -2, the last row becomes a row of zeros, and one parameter is needed to describe the solutions. That is, the system has infinitely many solutions.
 

Question 9

 
 
For which values of k will the system
x + y + z = 5
2x - 3y + z = 1
kx + 2y + kz = 3k
have a unique solution?
a) All k except k = 2.   b) All values of k
c) k = 2   d) k = -2 and 2.

 

Your answer is correct.
The augmented matrix
⌊ | ⌋ 1 1 1 |5 ⌈ 2 - 3 1 |1 ⌉ k 2 k 3k
reduces (after the application of a number of row operations) to
⌊ 1 1 1 | 5 ⌋ ⌈ 0 1 1 | 9 ⌉. 0 0 k5-2 |- 5k+18 5 | 5
The leading entry in row 3 is a non-zero number if and only if k⁄=2. Assuming that k⁄=2, we have three non-zero leading entries in rows 1, 2 and 3, leading to a unique solution.
Not correct. Choice (b) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations, working towards row echelon form. Then search for values of k which guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column 3) positions.
Not correct. Choice (c) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations, working towards row echelon form. Then search for values of k which guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column 3) positions.
Not correct. Choice (d) is false.
Hint: write down the augmented matrix of the system and apply a number of row operations, working towards row echelon form. Then search for values of k which guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column 3) positions.
 

Question 10

 
 
Select all the true statements from the options below.
a) Every linear system has at least one solution.   b) A linear system with 2 equations in 3 unknowns always has infinitely many solutions.
c) A linear system with more equations than unknowns is always inconsistent.   d) Every matrix is row equivalent to a unique reduced row echelon matrix.
e) Every row echelon matrix is also a reduced row echelon matrix.   f) Every linear system has either 0, 1 or infinitely many solutions.
g) There are five different reduced row echelon matrices with 2 rows and 3 columns, in which each entry is either 0 or 1.   h) A matrix can be row equivalent to more than one row echelon matrix.

 

There is at least one mistake.
For example, choice (a) should be false.
The system could be inconsistent (that is, have no solution).
There is at least one mistake.
For example, choice (b) should be false.
The system
x + y + z = 4
2x + 2y + 2z = 9
has 2 equations and 3 unknowns but is clearly inconsistent.
There is at least one mistake.
For example, choice (c) should be false.
The system
x + y = 4
2x + 2y = 8
5x + 5y = 20
has more equations than unknowns but has infinitely many solutions.
There is at least one mistake.
For example, choice (d) should be true.
There is at least one mistake.
For example, choice (e) should be false.
The matrix [1 2] 0 1 is in row echelon form but not reduced row echelon form.
There is at least one mistake.
For example, choice (f) should be true.
There is at least one mistake.
For example, choice (g) should be false.
In fact, there are fifteen such matrices. You can find them systematically by first counting how many have two leading 1’s (there are seven of these), then how many have one leading 1 (there are also seven of these) and then no leading 1 (there is one of these, the zero matrix). Of the seven with two leading 1s, four have the form [1 0 a] 0 1 b where a and b can be either 0 or 1, two have the form [1 a 0] 0 0 1 where a can be either 0 or 1, and the last is [0 1 0] 0 0 1. You can verify in a similar way that there are seven with one leading 1.
There is at least one mistake.
For example, choice (h) should be true.
Your answers are correct
  1. False. The system could be inconsistent (that is, have no solution).
  2. False. The system
    x + y + z = 4
    2x + 2y + 2z = 9
    has 2 equations and 3 unknowns but is clearly inconsistent.
  3. False. The system
    x + y = 4
    2x + 2y = 8
    5x + 5y = 20
    has more equations than unknowns but has infinitely many solutions.
  4. True.
  5. False. The matrix [1 2] 0 1 is in row echelon form but not reduced row echelon form.
  6. True.
  7. False. In fact, there are fifteen such matrices. You can find them systematically by first counting how many have two leading 1’s (there are seven of these), then how many have one leading 1 (there are also seven of these) and then no leading 1 (there is one of these, the zero matrix). Of the seven with two leading 1s, four have the form [1 0 a] 0 1 b where a and b can be either 0 or 1, two have the form [1 a 0] 0 0 1 where a can be either 0 or 1, and the last is [0 1 0] 0 0 1. You can verify in a similar way that there are seven with one leading 1.
  8. True.