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Quiz 8: Inverses and elementary matrices

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Question 1

 
 
Let A = [ ] - 5 7 3 - 4. Which of the following statements are true:
a) A is singular   b) A is invertible
c) A is non-singular   d) A -1 does not exist
e) The inverse of A is [ ] 4 7 3 5

 

There is at least one mistake.
For example, choice (a) should be false.
A is singular if A- 1 does not exist
There is at least one mistake.
For example, choice (b) should be true.
A is invertible is the same as saying that A is non-singular, or that -1 A exists. We can see that A is invertable by aplying elementary row operations:
[ | ] [ | ] [ | ] - 5 7 |1 0 1 - 1|1 2 1 - 1| 1 2 [A ∣ I] = 3 - 4 0 1 → 3 - 4 0 1 → 0 - 1 - 3 - 5 .
There is at least one mistake.
For example, choice (c) should be true.
A is non-singular is the same as saying that A is invertible, or that A -1 exists.
There is at least one mistake.
For example, choice (d) should be false.
You can see that A is invertible using elementary row operations.
There is at least one mistake.
For example, choice (e) should be true.
You can check that this matrix is the inverse of A by showing that [ ] [ ] A 4 7 = I = 4 7A 3 5 2 3 5 .
Your answers are correct
  1. False. A is singular if A- 1 does not exist
  2. True. A is invertible is the same as saying that A is non-singular, or that -1 A exists. We can see that A is invertable by aplying elementary row operations:
    [ | ] [ | ] [ | ] - 5 7 |1 0 1 - 1|1 2 1 - 1| 1 2 [A ∣ I] = 3 - 4 0 1 → 3 - 4 0 1 → 0 - 1 - 3 - 5 .
  3. True. A is non-singular is the same as saying that A is invertible, or that A -1 exists.
  4. False. You can see that A is invertible using elementary row operations.
  5. True. You can check that this matrix is the inverse of A by showing that [ ] [ ] A 4 7 = I = 4 7A 3 5 2 3 5 .
 

Question 2

 
 
Let A = ⌊ ⌋ ⌈- 3 - 1 - 2⌉ 2 3 4 1 4 5. The inverse of A (i.e., A-1) is:
a) undefined   b) B = ⌊- 3 - 1 - 2⌋ ⌈ 2 3 4⌉ 1 4 5
c) C = ⌊ 1 1⌋ |- 3 1 - 2| || || || 1 1 1 || || 2 3 4 || |⌈ 1 |⌉ 1 - 15 4   d) D = ⌊1 - 6 17⌋ ⌈4 - 2 1 ⌉ 11 - 8 0
e) none of the above

 

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.
We could check by carrying out matrix multiplication that none of AB, AC, AD is I, the 3 by 3 identity matrix, so (2), (3) and (4) are all incorrect. (In particular, note that A-1 does not consist of the inverses of the entries of A!!) This leaves (1) and (5), and to decide which of them is true, we can augment A by the 3 by 3 identity and reduce:
[A∣I] = ⌊ ⌋ - 3 - 1 - 2 1 0 0 ⌈ 2 3 4 0 1 0⌉ 1 4 5 0 0 1
→⌊ 1 4 5 0 0 1⌋ ⌈ 2 3 4 0 1 0⌉ - 3 - 1 - 2 1 0 0
→⌊ ⌋ 1 4 5 0 0 1 ⌈0 - 5 - 6 0 2 - 2⌉ 0 11 13 1 0 3
→⌊ ⌋ 1 4 5 0 0 1 ⌈0 1 6∕5 0 - 2∕5 2∕5⌉ 0 11 13 1 0 3 →⌊ ⌋ 1 4 5 0 0 1 ⌈ 0 1 6∕5 0 - 2∕5 2∕5 ⌉ 0 0 1∕5 1 22∕5 - 7∕5
at which point it is clear that there will be a leading 1 in row 3, column 3, so A-1 exists. Hence (5) is correct. We do not need to find A-1, but continuing the row reduction yields [I∣A-1], and in fact shows that
⌊ 1 3 - 2⌋ A- 1 = ⌈ 6 13 - 8⌉. - 5 - 11 7
 

Question 3

 
 
Let A = ⌊ ⌋ ⌈1 5 - 7⌉ 2 3 6 1 12 - 27. The inverse of A (i.e., A-1) is:
a) undefined   b) B = ⌊ ⌋ ⌈1 5 - 7⌉ 2 3 6 1 12 - 27
c) C = ⌊ 1 ⌋ |1 - - 17| || 5 || ||1 1 1 || ||2 3 6 || |⌈ |⌉ 1 -1 - -1 12 27   d) D = ⌊ ⌋ ⌈1 - 6 17⌉ 4 - 2 1 11 - 8 0
e) none of the above

 

Your answer is correct.
We could check by carrying out matrix multiplication that none of AB, AC, AD is I, the 3 by 3 identity matrix, so (2), (3) and (4) are all incorrect. (In particular, note that A-1 does not consist of the inverses of the entries of A!!) This leaves (1) and (5), and to decide which of them is true, we can augment A by the 3 by 3 identity and reduce:
[A∣I] = ⌊ ⌋ ⌈1 5 - 7 1 0 0⌉ 2 3 6 0 1 0 1 12 - 27 0 0 1
→⌊ ⌋ 1 5 - 7 1 0 0 ⌈0 - 7 20 - 2 1 0⌉ 0 7 - 20 - 1 0 1
→⌊1 5 - 7 1 0 0⌋ ⌈0 - 7 20 - 2 1 0⌉ 0 0 0 - 3 1 1
at which point it is clear that there can be no leading 1 in row 3, column 3, so (1) is correct.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 4

 
 
Which of the following matrices are elementary matrices?
a) ⌊ ⌋ 1 0 0 ⌈0 0 1⌉ 0 1 0   b) ⌊ ⌋ ⌈1 0 0⌉ 0 2 1 0 0 1
c) ⌊ 0 1 0⌋ ⌈ 0 0 1⌉ 1 0 0   d) ⌊ 1 2 0⌋ ⌈ 0 1 0⌉ 0 0 1

 

There is at least one mistake.
For example, choice (a) should be true.
This matrix corresponds to the elementary row operation R2 ↔ R3 .
There is at least one mistake.
For example, choice (b) should be false.
This matrix is the product of two elementary matrices corresponding to R2 := 2R2 then R2 := R2 + R3
There is at least one mistake.
For example, choice (c) should be false.
This matrix is product of two elementary matrices corresponding to R1 ↔ R3 then R2 ↔ R1
There is at least one mistake.
For example, choice (d) should be true.
This matrix to the elementary row operation R1 := R1 + 2R2 .
Your answers are correct
  1. True. This matrix corresponds to the elementary row operation R2 ↔ R3 .
  2. False. This matrix is the product of two elementary matrices corresponding to R2 := 2R2 then R2 := R2 + R3
  3. False. This matrix is product of two elementary matrices corresponding to R1 ↔ R3 then R2 ↔ R1
  4. True. This matrix to the elementary row operation R1 := R1 + 2R2 .
 

Question 5

 
 
The elementary matrix corresponding to the Elementary Row Operation R3 = R3 - 5R2 on a matrix with 3 rows is:
a) ⌊ ⌋ 1 0 0 ⌈0 1 - 5⌉ 0 0 1   b) ⌊ ⌋ 1 0 0 ⌈0 1 5⌉ 0 0 1
c) ⌊ ⌋ ⌈1 0 0⌉ 0 1 0 0 - 5 1   d) ⌊1 0 0⌋ ⌈0 1 0⌉ 0 5 1
e) none of the above

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
The correct answer is (3), since that is the result of applying the give elementary row operation to the 3 by 3 identity matrix.
Note that
⌊ ⌋⌊ ⌋ ⌊ ⌋ 1 0 0 a b ⋅⋅⋅ a b ⋅⋅⋅ ⌈0 1 0⌉⌈c d ⋅⋅⋅⌉ = ⌈ c d ⋅⋅⋅⌉ 0 - 5 1 e f ⋅⋅⋅ e - 5c f - 5d ⋅⋅⋅
is the matrix we would get by applying the given elementary row operation to ⌊ ⌋ ⌈a b ⋅⋅⋅⌉ c d ⋅⋅⋅ e f ⋅⋅⋅.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 6

 
 
The elementary matrix corresponding to the Elementary Row Operation R3 = 1 R3 9 on a matrix with 4 rows is:
a) ⌊1 0 0⌋ ⌈0 1 0⌉ 0 0 9   b) ⌊ ⌋ 1 0 0 |⌈0 1 0|⌉ 0 0 1 9
c) ⌊ ⌋ 1 0 0 0 ||0 1 0 0|| ⌈0 0 9 0⌉ 0 0 0 1   d) ⌊ ⌋ | 1 0 0 0 | || 0 1 01 0 || ⌈ 0 0 - 0 ⌉ 0 0 90 1
e) none of the above

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
The correct answer is (4), since that is the result of applying the give elementary row operation to the 4 by 4 identity matrix.
Note that
⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 1 0 0 0 a b ⋅⋅⋅ a b ⋅⋅⋅ || 0 1 0 0|| ||c d ⋅⋅⋅|| = || c d ⋅⋅⋅|| ⌈ 0 0 19 0⌉ ⌈e f ⋅⋅⋅⌉ ⌈19e 19f ⋅⋅⋅⌉ 0 0 0 1 g h ⋅⋅⋅ g h ⋅⋅⋅
is the matrix we would get by applying the given elementary row operation to ⌊a b ⋅⋅⋅⌋ |c d ⋅⋅⋅| |⌈e f ⋅⋅⋅|⌉ g h ⋅⋅⋅.
Not correct. Choice (e) is false.
 

Question 7

 
 
Consider the six elementary matrices below:
E1 = ⌊ 1 0 0 0⌋ | 0 1 0 0| |⌈ 0 0 1 0|⌉ 0 2 0 1,
E2 = ⌊ 0 1 0 0⌋ | 1 0 0 0| |⌈ 0 0 1 0|⌉ 0 0 0 1,
E3 = ⌊ 1 0 0 0⌋ | 0 1 0 0| |⌈ 0 0 1 0|⌉ 0 - 2 0 1,
   
E4 = ⌊ 0 0 1 0⌋ | 0 1 0 0| |⌈ 1 0 0 0|⌉ 0 0 0 1,
E5 = ⌊ 1 - 2 0 0⌋ | 0 1 0 0| |⌈ 0 0 1 0|⌉ 0 0 0 1,
E6 = ⌊ 1 2 0 0⌋ | 0 1 0 0| |⌈ 0 0 1 0|⌉ 0 0 0 1.
Which of the following statements are correct?
a) The inverse of E1 is E2.   b) The inverse of E1 is E3.
c) The inverse of E2 is E2.   d) The inverse of E5 is E6.
e) The inverse of E4 is E4.   f) The inverse of E3 is E6.

 

There is at least one mistake.
For example, choice (a) should be false.
E1 corresponds to the row operation R4 = R4 + 2R2, whereas E2 corresponds to R1←→R2.
There is at least one mistake.
For example, choice (b) should be true.
E1 corresponds to the row operation R4 = R4 + 2R2 and E3 corresponds to the row operation R4 = R4 - 2R2.
There is at least one mistake.
For example, choice (c) should be true.
E2 corresponds to R1←→R2.
There is at least one mistake.
For example, choice (d) should be true.
E5 corresponds to the row operation R1 = R1 - 2R2, whereas E6 corresponds to R1 = R1 + 2R2.
There is at least one mistake.
For example, choice (e) should be true.
E4 corresponds to R1←→R3.
There is at least one mistake.
For example, choice (f) should be false.
E3 corresponds to the row operation R4 = R4 - 2R2, whereas E6 corresponds to R1 = R1 + 2R2.
Your answers are correct
  1. False. E1 corresponds to the row operation R4 = R4 + 2R2, whereas E2 corresponds to R1←→R2.
  2. True. E1 corresponds to the row operation R4 = R4 + 2R2 and E3 corresponds to the row operation R4 = R4 - 2R2.
  3. True. E2 corresponds to R1←→R2.
  4. True. E5 corresponds to the row operation R1 = R1 - 2R2, whereas E6 corresponds to R1 = R1 + 2R2.
  5. True. E4 corresponds to R1←→R3.
  6. False. E3 corresponds to the row operation R4 = R4 - 2R2, whereas E6 corresponds to R1 = R1 + 2R2.
 

Question 8

 
 
Let A = ⌊3 - 1 1⌋ ⌈1 2 3⌉ 0 1 1, then A-1 = ⌊ 1 - 2 5 ⌋ ⌈ 1 - 3 6 ⌉ - 1 3 - 7.
What sequence of elementary row operations is needed to transform [A ∣I] to [I∣A -1] ?
a) R2 := R2 -1 3R1, R2 := 3R2, R2 ↔ R3, R2 := R2 - R3,
R3 := R3 - 7R2, R1 := R1 + R2, R1 := R1 - R3, R1 := 1 3R1
b) R1 := 1 3R1, R2 := R2 - R1, R2 := 3 7R2, R3 := R3 - R2
R3 := -7R3, R2 := R2 -8 7R3, R1 := R1 + 1 3R2, R1 := R1 -1 3R3
c) R2 := R2 -1 3R1, R3 := R3 - R2, R2 := 37R2, R3 := -7R3,
R2 := R1 - R3, R1 := R1 + R2, R1 := R1 - R3
d) All of the above sequences

 

Not correct. Choice (a) is false.
Your answer is correct.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 9

 
 
The system of equations
3x- y+ z = 1, x+ 2y+ 3z = 1, y+ z = - 1,
can be written in the form Ax = b where
⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 3 - 1 1 x 1 A = ⌈ 1 2 3⌉,x = ⌈y⌉ and b = ⌈ 1 ⌉. 0 1 1 z - 1
Using A-1 (given in question 8) to solve the system of equations, which of the following statements is correct ?
a) y = -8   b) z = -5
c) x = 6   d) None of the above

 

Your answer is correct.
A-1b = ⌊ ⌋ - 6 ⌈- 8⌉ 9 ⇒ x = -6, y = -8, z = 9.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 10

 
 
Let A = ⌊ ⌋ a b c ⌈d e f⌉ g h i, and suppose A-1 = ⌊ ⌋ 1 3 2 ⌈0 3 5 ⌉ 1 6 - 8. Consider the system
ax + by + cz = 5
dx + ey + fz = 1
gx + hy + iz = 4
Which of the following is true ?
a) the system has no solutions;   b) the system has many solutions;
c) the system has the unique solution x = 16, y = 23, z = -21;   d) the system has the unique solution x = 14, y = -22, z = 13;
e) none of the above.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Since A is invertible, there is the unique solution
⌊ ⌋ ⌈x⌉ y z = A-1⌊ ⌋ ⌈5⌉ 1 4
= ⌊ ⌋ 1 3 2 ⌈ 0 3 5 ⌉ 1 6 - 8⌊ ⌋ 5 ⌈1⌉ 4
= ⌊ 16 ⌋ ⌈ 23 ⌉ - 21.
Hence the correct answer is (3).
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.