Quiz 8: Inverses and elementary matrices

Question

Question 1

Let

A = [\begin{matrix} - 5 & 7 \\ 3 & - 4 \end{matrix}]

. Which of the following statements are true:

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

A^{- 1}

exists. We can see that

A

is invertable by aplying elementary row operations:

[A ∣ I] = [\begin{matrix} - 5 & 7 & 1 & 0 \\ 3 & - 4 & 0 & 1 \end{matrix}] \to [\begin{matrix} 1 & - 1 & 1 & 2 \\ 3 & - 4 & 0 & 1 \end{matrix}] \to [\begin{matrix} 1 & - 1 & 1 & 2 \\ 0 & - 1 & - 3 & - 5 \end{matrix}] .

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 [\begin{matrix} 4 & 7 \\ 3 & 5 \end{matrix}] = I_{2} = [\begin{matrix} 4 & 7 \\ 3 & 5 \end{matrix}] A

Your answers are correct

False. $A$ is singular if $A^{- 1}$ does not exist
True. $A$ is invertible is the same as saying that $A$ is non-singular, or that $A^{- 1}$ exists. We can see that $A$ is invertable by aplying elementary row operations: $[A ∣ I] = [\begin{matrix} - 5 & 7 & 1 & 0 \\ 3 & - 4 & 0 & 1 \end{matrix}] \to [\begin{matrix} 1 & - 1 & 1 & 2 \\ 3 & - 4 & 0 & 1 \end{matrix}] \to [\begin{matrix} 1 & - 1 & 1 & 2 \\ 0 & - 1 & - 3 & - 5 \end{matrix}] .$
True. $A$ is non-singular is the same as saying that $A$ is invertible, or that $A^{- 1}$ exists.
False. You can see that $A$ is invertible using elementary row operations.
True. You can check that this matrix is the inverse of $A$ by showing that $A [\begin{matrix} 4 & 7 \\ 3 & 5 \end{matrix}] = I_{2} = [\begin{matrix} 4 & 7 \\ 3 & 5 \end{matrix}] A$ .

Question 2

Let

A = [\begin{matrix} - 3 & - 1 & - 2 \\ 2 & 3 & 4 \\ 1 & 4 & 5 \end{matrix}]

. The inverse of

A

(i.e.,

A^{- 1}

) is:

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

A B

A C

A D

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:

\begin{array}{l} [A | I] & = [\begin{matrix} - 3 & - 1 & - 2 & 1 & 0 & 0 \\ 2 & 3 & 4 & 0 & 1 & 0 \\ 1 & 4 & 5 & 0 & 0 & 1 \end{matrix}] \\ \to [\begin{matrix} 1 & 4 & 5 & 0 & 0 & 1 \\ 2 & 3 & 4 & 0 & 1 & 0 \\ - 3 & - 1 & - 2 & 1 & 0 & 0 \end{matrix}] \\ \to [\begin{matrix} 1 & 4 & 5 & 0 & 0 & 1 \\ 0 & - 5 & - 6 & 0 & 2 & - 2 \\ 0 & 11 & 13 & 1 & 0 & 3 \end{matrix}] \\ \to [\begin{matrix} 1 & 4 & 5 & 0 & 0 & 1 \\ 0 & 1 & 6 ∕ 5 & 0 & - 2 ∕ 5 & 2 ∕ 5 \\ 0 & 11 & 13 & 1 & 0 & 3 \end{matrix}] \to [\begin{matrix} 1 & 4 & 5 & 0 & 0 & 1 \\ 0 & 1 & 6 ∕ 5 & 0 & - 2 ∕ 5 & 2 ∕ 5 \\ 0 & 0 & 1 ∕ 5 & 1 & 22 ∕ 5 & - 7 ∕ 5 \end{matrix}] \end{array}

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

A^{- 1} = [\begin{matrix} 1 & 3 & - 2 \\ 6 & 13 & - 8 \\ - 5 & - 11 & 7 \end{matrix}] .

Question 3

Let

A = [\begin{matrix} 1 & 5 & - 7 \\ 2 & 3 & 6 \\ 1 & 12 & - 27 \end{matrix}]

. The inverse of

A

(i.e.,

A^{- 1}

) is:

Your answer is correct.

We could check by carrying out matrix multiplication that none of

A B

A C

A D

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:

\begin{array}{l} [A | I] & = [\begin{matrix} 1 & 5 & - 7 & 1 & 0 & 0 \\ 2 & 3 & 6 & 0 & 1 & 0 \\ 1 & 12 & - 27 & 0 & 0 & 1 \end{matrix}] \\ \to [\begin{matrix} 1 & 5 & - 7 & 1 & 0 & 0 \\ 0 & - 7 & 20 & - 2 & 1 & 0 \\ 0 & 7 & - 20 & - 1 & 0 & 1 \end{matrix}] \\ \to [\begin{matrix} 1 & 5 & - 7 & 1 & 0 & 0 \\ 0 & - 7 & 20 & - 2 & 1 & 0 \\ 0 & 0 & 0 & - 3 & 1 & 1 \end{matrix}] \end{array}

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?

There is at least one mistake.
For example, choice (a) should be true.

This matrix corresponds to the elementary row operation

R_{2} \leftrightarrow R_{3}

There is at least one mistake.
For example, choice (b) should be false.

This matrix is the product of two elementary matrices corresponding to

R_{2} : = 2 R_{2}

then

R_{2} : = R_{2} + R_{3}

There is at least one mistake.
For example, choice (c) should be false.

This matrix is product of two elementary matrices corresponding to

R_{1} \leftrightarrow R_{3}

then

R_{2} \leftrightarrow R_{1}

There is at least one mistake.
For example, choice (d) should be true.

This matrix to the elementary row operation

R_{1} : = R_{1} + 2 R_{2}

Your answers are correct

True. This matrix corresponds to the elementary row operation $R_{2} \leftrightarrow R_{3}$ .
False. This matrix is the product of two elementary matrices corresponding to $R_{2} : = 2 R_{2}$ then $R_{2} : = R_{2} + R_{3}$
False. This matrix is product of two elementary matrices corresponding to $R_{1} \leftrightarrow R_{3}$ then $R_{2} \leftrightarrow R_{1}$
True. This matrix to the elementary row operation $R_{1} : = R_{1} + 2 R_{2}$ .

Question 5

The elementary matrix corresponding to the Elementary Row Operation

R_{3} = R_{3} - 5 R_{2}

on a matrix with 3 rows is:

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

[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 5 & 1 \end{matrix}] [\begin{matrix} a & b & \dots \\ c & d & \dots \\ e & f & \dots \end{matrix}] = [\begin{matrix} a & b & \dots \\ c & d & \dots \\ e - 5 c & f - 5 d & \dots \end{matrix}]

is the matrix we would get by applying the given elementary row operation to

[\begin{matrix} a & b & \dots \\ c & d & \dots \\ e & f & \dots \end{matrix}]

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

Question 6

The elementary matrix corresponding to the Elementary Row Operation

R_{3} = \frac{1}{9} R_{3}

on a matrix with 4 rows is:

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

[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{9} & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} a & b & \dots \\ c & d & \dots \\ e & f & \dots \\ g & h & \dots \end{matrix}] = [\begin{matrix} a & b & \dots \\ c & d & \dots \\ \frac{1}{9} e & \frac{1}{9} f & \dots \\ g & h & \dots \end{matrix}]

is the matrix we would get by applying the given elementary row operation to

[\begin{matrix} a & b & \dots \\ c & d & \dots \\ e & f & \dots \\ g & h & \dots \end{matrix}]

Not correct. Choice (e) is false.

Question 7

Consider the six elementary matrices below:

\begin{matrix} E_{1} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \end{matrix}], & E_{2} = [\begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}], & E_{3} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 2 & 0 & 1 \end{matrix}], \\ E_{4} = [\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}], & E_{5} = [\begin{matrix} 1 & - 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}], & E_{6} = [\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] . \end{matrix}

Which of the following statements are correct?

There is at least one mistake.
For example, choice (a) should be false.

E_{1}

corresponds to the row operation

R_{4} = R_{4} + 2 R_{2}

, whereas

E_{2}

corresponds to

R_{1} \leftrightarrow R_{2}

There is at least one mistake.
For example, choice (b) should be true.

E_{1}

corresponds to the row operation

R_{4} = R_{4} + 2 R_{2}

and

E_{3}

corresponds to the row operation

R_{4} = R_{4} - 2 R_{2}

There is at least one mistake.
For example, choice (c) should be true.

E_{2}

corresponds to

R_{1} \leftrightarrow R_{2}

There is at least one mistake.
For example, choice (d) should be true.

E_{5}

corresponds to the row operation

R_{1} = R_{1} - 2 R_{2}

, whereas

E_{6}

corresponds to

R_{1} = R_{1} + 2 R_{2}

There is at least one mistake.
For example, choice (e) should be true.

E_{4}

corresponds to

R_{1} \leftrightarrow R_{3}

There is at least one mistake.
For example, choice (f) should be false.

E_{3}

corresponds to the row operation

R_{4} = R_{4} - 2 R_{2}

, whereas

E_{6}

corresponds to

R_{1} = R_{1} + 2 R_{2}

Your answers are correct

False. $E_{1}$ corresponds to the row operation $R_{4} = R_{4} + 2 R_{2}$ , whereas $E_{2}$ corresponds to $R_{1} \leftrightarrow R_{2}$ .
True. $E_{1}$ corresponds to the row operation $R_{4} = R_{4} + 2 R_{2}$ and $E_{3}$ corresponds to the row operation $R_{4} = R_{4} - 2 R_{2}$ .
True. $E_{2}$ corresponds to $R_{1} \leftrightarrow R_{2}$ .
True. $E_{5}$ corresponds to the row operation $R_{1} = R_{1} - 2 R_{2}$ , whereas $E_{6}$ corresponds to $R_{1} = R_{1} + 2 R_{2}$ .
True. $E_{4}$ corresponds to $R_{1} \leftrightarrow R_{3}$ .
False. $E_{3}$ corresponds to the row operation $R_{4} = R_{4} - 2 R_{2}$ , whereas $E_{6}$ corresponds to $R_{1} = R_{1} + 2 R_{2}$ .

Question 8

Let

A = [\begin{matrix} 3 & - 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 1 \end{matrix}]

, then

A^{- 1} = [\begin{matrix} 1 & - 2 & 5 \\ 1 & - 3 & 8 \\ - 1 & 3 & - 7 \end{matrix}]

.
What sequence of elementary row operations is needed to transform

[A | I]

[I | A^{- 1}]

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

\begin{array}{rcl} 3 x - y + z & = & 1, \\ x + 2 y + 3 z & = & 1, \\ y + z & = & - 1, \end{array}

can be written in the form

A x = b

where

A = [\begin{matrix} 3 & - 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 1 \end{matrix}], x = [\begin{matrix} x \\ y \\ z \end{matrix}] a n d b = [\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}] .

Using

A^{- 1}

(given in question 8) to solve the system of equations, which of the following statements is correct ?

Your answer is correct.

A^{- 1} b = [\begin{matrix} - 6 \\ - 10 \\ 9 \end{matrix}] \Rightarrow x = - 6

y = - 10

z = 9

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Question 10

Let

A = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}]

, and suppose

A^{- 1} = [\begin{matrix} 1 & 3 & 2 \\ 0 & 3 & 5 \\ 1 & 6 & - 8 \end{matrix}]

. Consider the system

\begin{array}{l} a x + b y + c z & = 5 \\ d x + e y + f z & = 1 \\ g x + h y + i z & = 4 \end{array}

Which of the following is true ?

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

\begin{array}{l} [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} [\begin{matrix} 5 \\ 1 \\ 4 \end{matrix}] \\ = [\begin{matrix} 1 & 3 & 2 \\ 0 & 3 & 5 \\ 1 & 6 & - 8 \end{matrix}] [\begin{matrix} 5 \\ 1 \\ 4 \end{matrix}] \\ = [\begin{matrix} 16 \\ 23 \\ - 21 \end{matrix}] . \end{array}

Hence the correct answer is (3).

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

Questions:

right first attempt
right
wrong

The University of Sydney - School of Mathematics and Statistics

Quiz 8: Inverses and elementary matrices

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Sitemap

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 $[\begin{matrix} 4 & 7 \\ 3 & 5 \end{matrix}]$

a)	undefined	b)	$B = [\begin{matrix} - 3 & - 1 & - 2 \\ 2 & 3 & 4 \\ 1 & 4 & 5 \end{matrix}]$
c)	$C = [\begin{matrix} - \frac{1}{3} & 1 & - \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ 1 & \frac{1}{4} & \frac{1}{5} \end{matrix}]$	d)	$D = [\begin{matrix} 1 & - 6 & 17 \\ 4 & - 2 & 1 \\ 11 & - 8 & 0 \end{matrix}]$
e)	none of the above

a)	undefined	b)	$B = [\begin{matrix} 1 & 5 & - 7 \\ 2 & 3 & 6 \\ 1 & 12 & - 27 \end{matrix}]$
c)	$C = [\begin{matrix} 1 & \frac{1}{5} & - \frac{1}{7} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ 1 & \frac{1}{12} & - \frac{1}{27} \end{matrix}]$	d)	$D = [\begin{matrix} 1 & - 6 & 17 \\ 4 & - 2 & 1 \\ 11 & - 8 & 0 \end{matrix}]$
e)	none of the above

a)		$[\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$		b)		$[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}]$
c)		$[\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$		d)		$[\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

a)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - 5 \\ 0 & 0 & 1 \end{matrix}]$	b)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix}]$
c)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 5 & 1 \end{matrix}]$	d)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 1 \end{matrix}]$
e)	none of the above

a)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{matrix}]$	b)	$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{9} \end{matrix}]$
c)	$[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 9 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]$	d)	$[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{9} & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]$
e)	none of the above

a)	The inverse of $E_{1}$ is $E_{2}$ .	b)	The inverse of $E_{1}$ is $E_{3}$ .
c)	The inverse of $E_{2}$ is $E_{2}$ .	d)	The inverse of $E_{5}$ is $E_{6}$ .
e)	The inverse of $E_{4}$ is $E_{4}$ .	f)	The inverse of $E_{3}$ is $E_{6}$ .

a)		$R_{2} : = R_{2} - \frac{1}{3} R_{1}$ , $R_{2} : = 3 R_{2}$ , $R_{2} \leftrightarrow R_{3}$ , $R_{2} : = R_{2} - R_{3},$ $R_{3} : = R_{3} - 7 R_{2}$ , $R_{1} : = R_{1} + R_{2}$ , $R_{1} : = R_{1} - R_{3}$ , $R_{1} : = \frac{1}{3} R_{1}$
b)		$R_{1} : = \frac{1}{3} R_{1}$ , $R_{2} : = R_{2} - R_{1}$ , $R_{2} : = \frac{3}{7} R_{2}$ , $R_{3} : = R_{3} - R_{2}$ $R_{3} : = - 7 R_{3}$ , $R_{2} : = R_{2} - \frac{8}{7} R_{3}$ , $R_{1} : = R_{1} + \frac{1}{3} R_{2}$ , $R_{1} : = R_{1} - \frac{1}{3} R_{3}$
c)		$R_{2} : = R_{2} - \frac{1}{3} R_{1}$ , $R_{3} : = R_{3} - R_{2}$ , $R_{2} : = \frac{3}{7} R_{2}$ , $R_{3} : = - 7 R_{3},$ $R_{2} : = R_{1} - R_{3}$ , $R_{1} : = R_{1} + R_{2}$ , $R_{1} : = R_{1} - R_{3}$
d)		All of the above sequences

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.

a)	$y = - 10$	b)	$z = - 5$
c)	$x = 6$	d)	None of the above