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MATH1002 Quizzes

Quiz 9: Inverses and elementary row operations
Question 1 Questions
A sequence of elementary row operations transforms the augmented matrix A|I into
130123 0 1 2 1 0 2 001231 .
Find A1. Exactly one option must be correct)
a)
130 0 1 2 001
b)
102033 6 0 2 3 1
c)
123 1 0 2 231
d)
211 3 3 2 2 3 1

Choice (a) is incorrect
Choice (b) is correct!
The two operations that need to be performed are R2 := R2 2R3 and R1 := R1 3R2.
Choice (c) is incorrect
You still need to perform two operations in order to reduce the left hand matrix to the identity matrix.
Choice (d) is incorrect
Suppose a sequence of elementary row operations has been performed on A|I and has resulted in
1 1 22 1 23 0 1 0 1020 0 12 11 .
Find A1. Exactly one option must be correct)
a)
5 200 1 02 2 1 1
b)
3 2 2 2 520 2 1 1
c)
7 202 1 02 2 1 1
d)
1 224 3 24 2 1 1

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The two elementary row operations that need to be performed are R1 := R1 + 1 2R2 and R1 := R1 2R3.
Choice (d) is incorrect
Suppose A is a 9 × 9 matrix that can be reduced to to the 9 × 9 identity matrix by applying four elementary row operations. Let E1, E2, E3, and E4, respectively, be the elementary matrices corresponding to these four row operations. Which of the following is true ? Exactly one option must be correct)
a)
A = E1E2E3E4
b)
A = E4E3E2E1
c)
A = E41E31E21E11
d)
A = E11E21E31E41
e)
None of the above

Choice (a) is incorrect
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.
Choice (b) is incorrect
The matrix E4E3E2E1 is the inverse of A.
Choice (c) is incorrect
We have that E4E3E2E1A = I, where I is the 9 × 9 identity matrix.
Choice (d) 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 E41, and then by E31, E21 and E11 gives A = E41E31E21E11 as claimed.
Choice (e) is incorrect
Recall that performing an elementary row operation is equivalent to multiplying on the left by the corresponding elementary matrix.
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. Exactly one option must be correct)
a)
100 0 1 3 001
b)
10 0 0 1 3 00 1
c)
1000 0 1 3 0 0010 0 0 0 1
d)
10 0 0 0 1 3 0 00 1 0 0 0 0 1

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Since A is transformed to I4, A must be a 4 × 4 matrix and the elementary matrices are also 4 × 4. Now
E2 = 1000 0 1 3 0 0010 0 0 0 1 andF2 = E21 = 10 0 0 0 1 3 0 00 1 0 0 0 0 1
.
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. Exactly one option must be correct)
a)
120 0 1 0 0 0 1
b)
120 0 1 0 001
c)
100 2 1 0 001
d)
1 002 1 0 0 01

Choice (a) is correct!
F4 = E41 so E4 = F41.
F4 = 120 0 1 0 001 hence E4 = 120 0 1 0 0 0 1 .
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
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 ? Exactly one option must be correct)
a)
A = 13 2 5
b)
A1 = 13 2 5
c)
A = 1 3 2 5
d)
A1 = 1 3 2 5

Choice (a) is correct!
E1 = 1 02 1 , E2 = 13 0 1 .
A = E11E21 = 10 2 1 13 0 1 = 13 2 5 .
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
A certain matrix A can be transformed to the 2 × 2 identity matrix I2 by the following sequence of elementary row operations:
R1 := 1 2R1,R2 := R2 R1,R2 = 1 4R2,R1 := R1 2R2.
Find the matrices A and A1. (Zero or more options can be correct)
a)
A = 3 4 1 20 1 8 1 4 0 0 0 1
b)
A1 = 240 1 6 0 001
c)
A = 2 001 6 0 0 01
d)
A1 = 1 2 00 1 121 60 0 0 1
e)
A = 24 1 6
f)
A1 = 3 4 1 2 1 8 1 4
g)
A = 1 2 0 1 121 6
h)
A1 = 2 01 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.
E1 = 1 20 0 1 , E2 = 1 01 1 , E3 = 10 01 4 , E4 = 12 0 1 . So A = E11E21E31E41 = 24 1 6 and A1 = E 4E3E2E1 = 3 4 1 2 1 8 1 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.
Correct!
  1. False
  2. False
  3. False
  4. False
  5. True
  6. True E1 = 1 20 0 1 , E2 = 1 01 1 , E3 = 10 01 4 , E4 = 12 0 1 . So A = E11E21E31E41 = 24 1 6 and A1 = E 4E3E2E1 = 3 4 1 2 1 8 1 4 .
  7. False
  8. False
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? Exactly one option must be correct)
a)
A = 31 3 17 6 11 3 10 3 4 3 2 3 2 1 2 1
b)
A = 1 1 3 2 3 16 3 11 4
c)
A = 35 67 3 24 3 2 3 01 2 1
d)
A = 1 1 3 2 3 20 3 1 3 2 13 3

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The elementary matrices corresponding to the above elementary row operations are, in the order of application,
E1 = 100 2 1 0 001 , E2 = 1 00 0 1 0 101 , E3 = 100 0 1 0 01 21 ,
E4 = 103 0 1 0 00 1 , E5 = 100 012 3 001 andE6 = 110 0 1 0 001 .
A = E11E21E31E41E51E61 = 1 1 3 2 3 20 3 1 3 2 13 3 .
Suppose that A = 1 002 1 0 0 01 100 0 1 0 301 100 0 3 0 001 1 0 0 0 1 0 021 10 0 0 1 3 00 1 . Find A1. Exactly one option must be correct)
a)
A1 = 100 01 30 0 0 1 100 0 1 0 021 100 0 1 3 001 102 0 1 0 00 1 110 0 1 0 001
b)
A1 = 1 0 0 0 3 0 0 0 1 100 0 1 0 021 100 0 1 3 001 102 0 1 0 00 1 110 0 1 0 001
c)
A1 = 100 0 1 3 001 100 0 1 0 021 100 01 30 0 0 1 1 00 0 1 0 301 100 2 1 0 001
d)
A1 = 100 0 1 3 001 100 0 1 0 021 1 0 0 0 3 0 0 0 1 1 00 0 1 0 301 100 2 1 0 001

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
If A = E1E2E5 then A1 = E51E41E11
Choice (d) is incorrect
Suppose that A1 = 100 0 1 1 001 100 0 0 1 010 100 2 1 0 001 100 0 2 0 001 110 0 1 0 001 . Find A. Exactly one option must be correct)
a)
A = 100 0 0 1 010 1 002 1 0 0 01 1 0 0 0 2 0 0 0 1 110 0 1 0 0 0 1 102 0 1 0 001
b)
A = 110 0 1 0 0 0 1 100 01 20 0 0 1 1 002 1 0 0 01 100 0 0 1 010 10 0 0 1 1 00 1
c)
A = 110 0 1 0 0 0 1 1 0 0 0 2 0 0 0 1 1 002 1 0 0 01 001 0 1 0 100 10 0 0 1 1 00 1
d)
A = 100 0 0 1 010 1 002 1 0 0 01 100 01 20 0 0 1 110 0 1 0 0 0 1 102 0 1 0 001

Choice (a) is incorrect
Choice (b) is correct!
If A1 = E1E2E5 then A = E51E41E11
Choice (c) is incorrect
Choice (d) is incorrect