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Quiz 9: Inverses and elementary row operations

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Question 1

 
 
A sequence of elementary row operations transforms the augmented matrix [A ∣I] into
⌊ | ⌋ 1 3 0|1 2 3 ⌈ 0 1 2|1 0 2 ⌉ 0 0 1 2 3 1.
Find A-1.
a) ⌊ ⌋ 1 3 0 ||0 1 2|| ⌈0 0 1⌉   b) ⌊ ⌋ 10 20 3 ⌈- 3 - 6 0⌉ 2 3 1
c) ⌊1 2 3⌋ ⌈1 0 2⌉ 2 3 1   d) ⌊ ⌋ - 2 - 1 - 1 ⌈ 3 3 2 ⌉ 2 3 1

 

Not correct. Choice (a) is false.
Your answer is correct.
The two operations that need to be performed are R2 := R2 - 2R3 and R1 := R1 - 3R2 .
Not correct. Choice (c) is false.
You still need to perform two operations in order to reduce the left hand matrix to the identity matrix.
Not correct. Choice (d) is false.
 

Question 2

 
 
Suppose a sequence of elementary row operations has been peformed on [A∣I] and has resulted in
| ⌊ 1 | ⌋ | 1 -2 2| 1 2 3| . ⌈ 0 1 0|- 1 0 2⌉ 0 0 1 2 1 1
Find A-1.
a) ⌊ 5 ⌋ |- 2 0 0| ⌈- 1 0 2⌉ 2 1 1   b) ⌊ ⌋ 3 2 2 |⌈ 2- 5 - 2 0|⌉ 2 1 1
c) ⌊ 7 ⌋ |- 2 0 2| ⌈- 1 0 2⌉ 2 1 1   d) ⌊ 1 ⌋ | 2 2 4| ⌈ 3 2 4⌉ 2 1 1

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
The two elementary row operations that need to be performed are R1 := R1 + 1R2 2 and R1 := R1 - 2R3 .
Not correct. Choice (d) is false.
 

Question 3

 
 
Suppose A is a 9 × 9 and that which can be reduced to to the 9 × 9 identity matrix by applying four elementary row oprations. Let E1, E2, E3, and E4, respectively, be the elementary matrices corresponding to these four row operations. Which of the following is true ?
a) A = E1E2E3E4
b) A = E4E3E2E1
c) A = E4-1E3-1E2-1E1-1
d) A = E1-1E2-1E3-1E4-1
e) None of the above

 

Not correct. Choice (a) is false.
Multiplying A on the left by E1E2E3E4 would correspond to applying the fourth elementary row operation to A, followed by the third,m then second and then the first.
Not correct. Choice (b) is false.
The matrix E4E3E2E1 is the inverse of A.
Not correct. Choice (c) is false.
We have that E4E3E2E1A = I, where I is the 9 × 9 identity matrix.
Your answer is correct.
Recall that performing an elementary row operation is equivalent to multiplying on the left by the corresponding elementary matrix. So applying these four elementary operations to A in order gives the sequence of matrices:
A → E1A → E2E1A → E3E2E1A → E4E3E2E1A
Hence, E4E3E2E1A = I, where I is the 9 × 9 identity matrix. Multiplying on the left first by E4-1, and then by E3-1, E2-1 and E1-1 gives A = E4-1E3-1E2-1E1-1 as claimed.
Not correct. Choice (e) is false.
Recall that performing an elementary row operation is equivalent to multiplying on the left by the corresponding elementary matrix.
 

Question 4

 
 
Suppose that a matrix A can be transformed to I4 by a sequence of elementary row operations and let E1, E2, E3 and E4 be the elementary matrices which correspond to these elementary row operations, in the order in which they are applied. Then A = F1F2F3F4. If E2 corresponds to the elementary row operation R2 := R2 + 3R3, find F2.
a) ⌊ ⌋ ⌈1 0 0⌉ 0 1 3 0 0 1   b) ⌊1 0 0 ⌋ ⌈0 1 - 3⌉ 0 0 1
c) ⌊1 0 0 0⌋ |0 1 3 0| |⌈0 0 1 0|⌉ 0 0 0 1   d) ⌊1 0 0 0⌋ |0 1 - 3 0| |⌈0 0 1 0|⌉ 0 0 0 1

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
Since A is transformed to I4, A must be a 4 × 4 matrix and the elementary matrices are also 4 × 4. Now
⌊ ⌋ ⌊ ⌋ 1 0 0 0 1 0 0 0 E = ||0 1 3 0|| and F = E- 1= ||0 1 - 3 0|| 2 ⌈0 0 1 0⌉ 2 2 ⌈0 0 1 0⌉ 0 0 0 1 0 0 0 1
.
 

Question 5

 
 
Suppose the matrix A can be transformed to I3 by a sequence of elementary row operations and let E1, E2, E3 and E4 be the elementary matrices which correspond to these elementary row operations in the order in which they are applied. Then A = F1F2F3F4. If F4 corresponds to the elementary row operation R1 := R1 + 2R2, find E4.
a) ⌊ ⌋ 1 - 2 0 ⌈0 1 0⌉ 0 0 1   b) ⌊ ⌋ ⌈1 2 0⌉ 0 1 0 0 0 1
c) ⌊ ⌋ ⌈1 0 0⌉ 2 1 0 0 0 1   d) ⌊ 1 0 0⌋ ⌈- 2 1 0⌉ 0 0 1

 

Your answer is correct.
F4 = E4-1 so E4 = F4-1.
F4 = ⌊ ⌋ 1 2 0 ⌈ 0 1 0⌉ 0 0 1 hence E4 = ⌊ ⌋ 1 - 2 0 ⌈0 1 0⌉ 0 0 1.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 6

 
 
Suppose the matrix A can be transformed to I2 by the sequence of elementary row operations R2 := R2 - 2R1, R1 := R1 + 3R2. Which of the following statements is true ?
a) A = [ ] 1 - 3 2 - 5   b) A-1 = [ ] 1 - 3 2 - 5
c) A = [ ] 1 3 - 2 - 5   d) A-1 = [ ] 1 3 - 2 - 5

 

Your answer is correct.
E1 = [ ] 1 0 - 2 1, E2 = [ ] 1 3 0 1.
A = E1-1E2-1 = [ ] 1 0 2 1[ ] 1 - 3 0 1 = [ ] 1 - 3 2 - 5.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 7

 
 
A certain matrix A can be transformed to the 2 × 2 identgity matrix I2 by the following sequence of elementary row operations:
R := 1R ,R := R - R ,R = 1R ,R := R - 2R . 1 2 1 2 2 1 2 4 2 1 1 2
Find the matrices A and A-1.
a) ⌊ 3 1 ⌋ 41 -21 0 A = ⌈- 8 4 0⌉ 0 0 1   b) A-1 = ⌊ ⌋ ⌈2 4 0⌉ 1 6 0 0 0 1
c) A = ⌊ ⌋ ⌈ 2 0 0⌉ - 1 6 0 0 0 1   d) ⌊ 1 ⌋ -1 ⌈ 21- 01 0⌉ A = 12 6 0 0 0 1
e) A = [2 4] 1 6   f) -1 [34 - 12] A = - 18 14
g) [ ] 12 0 A = 112 16   h) A-1 = [ ] 2 0 - 1 6

 

There is at least one mistake.
For example, choice (a) should be false.
There is at least one mistake.
For example, choice (b) should be false.
There is at least one mistake.
For example, choice (c) should be false.
There is at least one mistake.
For example, choice (d) should be false.
There is at least one mistake.
For example, choice (e) should be true.
There is at least one mistake.
For example, choice (f) should be true.
[ ] 12 0 E1 = 0 1 , E2 = [ ] 1 0 - 1 1, [ ] 1 0 E3 = 0 14 , E4 = [ ] 1 - 2 0 1. So A = E1-1E2-1E3-1E4-1 = [2 4] 1 6 and -1 [3- - 1] A = E4E3E2E1 = -41 12 8 4 .
There is at least one mistake.
For example, choice (g) should be false.
There is at least one mistake.
For example, choice (h) should be false.
Your answers are correct
  1. False.
  2. False.
  3. False.
  4. False.
  5. True.
  6. True. [ ] 12 0 E1 = 0 1 , E2 = [ ] 1 0 - 1 1, [ ] 1 0 E3 = 0 14 , E4 = [ ] 1 - 2 0 1. So A = E1-1E2-1E3-1E4-1 = [2 4] 1 6 and -1 [3- - 1] A = E4E3E2E1 = -41 12 8 4 .
  7. False.
  8. False.
 

Question 8

 
 
A matrix A can be transformed to I3, the 3 × 3 identity matrixm by the following sequence of elementary row operations:
R2 := R2 + 2R1 R3 := R3 - R1
R3 := R3 + 1 2R2 R1 := R1 - 3R3
R2 := R2 + 2 3R3 R1 := R1 + R2.
Which of the following matrices is equal to A?
a) A = ⌊ ⌋ 313 - 176- 131 ⌈- 103 43 - 23⌉ 2 - 12 1   b) A = ⌊ 1 - 1 - 3 ⌋ ⌈ 2 3 - 163⌉ - 1 - 1 4
c) A = ⌊ ⌋ 3 56 - 73 ⌈2 43 23⌉ 0 12 1   d) A = ⌊ ⌋ 1 - 1 3 ⌈- 2 3 - 203⌉ 1 - 32 133-

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
The elementary matrices corresponding to the above elementary row operations are, in the order of application,
E1 = ⌊ ⌋ 1 0 0 ⌈ 2 1 0⌉ 0 0 1, E2 = ⌊ ⌋ 1 0 0 ⌈ 0 1 0⌉ - 1 0 1, E3 = ⌊ ⌋ 1 0 0 ⌈0 11 0⌉ 0 2 1,
E4 = ⌊ ⌋ 1 0 - 3 ⌈ 0 1 0 ⌉ 0 0 1, E5 = ⌊ ⌋ 1 0 0 ⌈0 1 23⌉ 0 0 1 and E6 = ⌊ ⌋ 1 1 0 ⌈0 1 0⌉ 0 0 1.
A = E1-1E2-1E3-1E4-1E5-1E6-1 = ⌊ ⌋ 1 - 1 3 ⌈- 2 3 - 203⌉ 1 - 32 133.
 

Question 9

 
 
Suppose that
⌊ 1 0 0⌋⌊ 1 0 0⌋ ⌊1 0 0⌋⌊ 1 0 0⌋⌊ 1 0 0 ⌋ A = ⌈ - 2 1 0⌉⌈ 0 1 0⌉ ⌈0 3 0⌉⌈ 0 1 0⌉⌈ 0 1 - 3⌉. 0 0 1 3 0 1 0 0 1 0 - 2 1 0 0 1
Find A-1.
a)
⌊1 0 0⌋ ⌊1 0 0⌋⌊ 1 0 0⌋ ⌊1 0 - 2⌋ ⌊1 1 0⌋ A -1 = ⌈0 1 0⌉ ⌈0 1 0⌉⌈ 0 1 3⌉ ⌈0 1 0⌉ ⌈0 1 0⌉ 0 30 1 0 2 1 0 0 1 0 0 1 0 0 1
b)
⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ -1 1 0 0 1 0 0 1 0 0 1 0 - 2 1 1 0 A = ⌈0 - 3 0⌉ ⌈0 1 0⌉ ⌈0 1 3⌉ ⌈0 1 0 ⌉ ⌈0 1 0⌉ 0 0 1 0 2 1 0 0 1 0 0 1 0 0 1
c)
⌊ ⌋ ⌊ ⌋⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 A -1 = ⌈0 1 3⌉ ⌈0 1 0⌉⌈0 13 0⌉ ⌈ 0 1 0⌉ ⌈2 1 0⌉ 0 0 1 0 2 1 0 0 1 - 3 0 1 0 0 1
d)
⌊1 0 0⌋ ⌊1 0 0⌋ ⌊1 0 0⌋ ⌊ 1 0 0⌋ ⌊1 0 0⌋ A -1 = ⌈0 1 3⌉ ⌈0 1 0⌉ ⌈0 - 3 0⌉ ⌈ 0 1 0⌉ ⌈2 1 0⌉ 0 0 1 0 2 1 0 0 1 - 3 0 1 0 0 1

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
If A = E1E2…E5 then A-1 = E5-1E4-1…E1-1
Not correct. Choice (d) is false.
 

Question 10

 
 
Suppose that
⌊ 1 0 0⌋ ⌊1 0 0⌋⌊ 1 0 0⌋ ⌊ 1 0 0⌋ ⌊1 1 0⌋ A -1 = ⌈ 0 1 1⌉ ⌈0 0 1⌉⌈ 2 1 0⌉ ⌈ 0 2 0⌉ ⌈0 1 0⌉. 0 0 1 0 1 0 0 0 1 0 0 1 0 0 1
Find A.
a)
⌊1 0 0⌋⌊ 1 0 0⌋⌊ 1 0 0⌋⌊ 1 - 1 0⌋⌊ 1 0 2⌋ A = ⌈0 0 1⌉⌈- 2 1 0⌉⌈ 0 - 2 0⌉⌈ 0 1 0⌉⌈ 0 1 0⌉ 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1
b)
⌊ ⌋⌊ ⌋ ⌊ ⌋ ⌊ ⌋⌊ ⌋ 1 - 1 0 1 0 0 1 0 0 1 0 0 1 0 0 A = ⌈ 0 1 0⌉⌈ 0 12 0⌉ ⌈- 2 1 0⌉ ⌈0 0 1⌉⌈0 1 - 1⌉ 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1
c)
⌊ ⌋ ⌊ ⌋ ⌊ ⌋⌊ ⌋ ⌊ ⌋ 1 - 1 0 1 0 0 1 0 0 0 0 1 1 0 0 A = ⌈0 1 0⌉ ⌈0 - 2 0⌉ ⌈- 2 1 0⌉⌈0 1 0⌉ ⌈0 1 - 1⌉ 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1
d)
⌊1 0 0⌋ ⌊ 1 0 0⌋ ⌊1 0 0⌋ ⌊1 - 1 0⌋ ⌊1 0 2⌋ A = ⌈0 0 1⌉ ⌈- 2 1 0⌉ ⌈0 1 0⌉ ⌈0 1 0⌉ ⌈0 1 0⌉ 0 1 0 0 0 1 0 20 1 0 0 1 0 0 1

 

Not correct. Choice (a) is false.
Your answer is correct.
If A-1 = E1E2…E5 then A = E5-1E4-1…E1-1
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.