Quiz 6: Reduced row echelon form

Question

Question 1

Which of the following matrices are in reduced row echelon form?

a)	$[\begin{matrix} 1 & 3 & 5 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]$	b)	$[\begin{matrix} 1 & 3 & 0 & 5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]$
c)	$[\begin{matrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]$	d)	$[\begin{matrix} 1 & 3 & 0 & 5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$
e)	$[\begin{matrix} 1 & 9 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]$

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

The leading 1 in row 2 is not the only non-zero entry in its column (column 3).

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

The leading 1 in row 3 also has a non-zero in its column (column 4).

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

This matrix contains a row of zeros with a non–zero row below it and, in addition, the leading 1 in row 4 has a non–zero entry in its column (column 4).

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

The entries above and below each leading 1 are all zero.

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

This matrix contains a row of zeros with a non–zero row below it.

Your answers are correct

False. The leading 1 in row 2 is not the only non-zero entry in its column (column 3).
False. The leading 1 in row 3 also has a non-zero in its column (column 4).
False. This matrix contains a row of zeros with a non–zero row below it and, in addition, the leading 1 in row 4 has a non–zero entry in its column (column 4).
True. The entries above and below each leading 1 are all zero.
False. This matrix contains a row of zeros with a non–zero row below it.

Question 2

Which of the following matrices are in reduced row echelon form?

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

This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.

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

This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.

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

This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.

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

This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.

Your answers are correct

True. This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.
True. This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.
True. This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.
True. This is in row echelon form and the entries above and below each leading 1 and in the same column all zero.

Question 3

Let

A = [\begin{matrix} 2 & 6 & 1 & - 7 \\ 1 & 3 & 1 & - 3 \\ 1 & 3 & 4 & 0 \end{matrix}]

Which of the following sequences of elementary row operations transforms

A

into row echelon form? (More than one answer may be correct.)

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

This sequence of operations transforms

A

into the matrix

[\begin{matrix} 1 & 3 & - 3 & - 7 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 7 & 7 \end{matrix}]

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

This sequence of operations transforms

A

into the matrix

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

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

This sequence of operations transforms

A

into the matrix

[\begin{matrix} 1 & 3 & \frac{1}{2} & - \frac{7}{2} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}] .

Your answers are correct

False. This sequence of operations transforms $A$ into the matrix $[\begin{matrix} 1 & 3 & - 3 & - 7 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 7 & 7 \end{matrix}]$
True. This sequence of operations transforms $A$ into the matrix $[\begin{matrix} 1 & 3 & 1 & - 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]$
True. This sequence of operations transforms $A$ into the matrix $[\begin{matrix} 1 & 3 & \frac{1}{2} & - \frac{7}{2} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}] .$

Question 4

What is the reduced row echelon form of the matrix

[\begin{matrix} 2 & 4 & 6 & - 8 \\ 3 & 6 & 7 & 10 \end{matrix}]

a)	$[\begin{matrix} 1 & 2 & 3 & - 4 \\ 1 & 2 & \frac{7}{3} & \frac{10}{3} \end{matrix}]$	b)	$[\begin{matrix} 1 & 0 & 3 & - 4 \\ 0 & 1 & - 2 & 22 \end{matrix}]$
c)	$[\begin{matrix} 1 & 2 & 0 & 29 \\ 0 & 0 & 1 & - 11 \end{matrix}]$	d)	$[\begin{matrix} 1 & 2 & 0 & 37 \\ 0 & 0 & 1 & - 11 \end{matrix}]$
e)	$[\begin{matrix} 1 & 2 & 3 & - 4 \\ 0 & 0 & 1 & - 11 \end{matrix}]$

Not correct. Choice (a) is false.

This matrix is not even is row echelon form!

Not correct. Choice (b) is false.

Check your sequence of row operations.

Your answer is correct.

The following sequence of elementary row operations:

R_{1} : = \frac{1}{2} R_{1}, R_{2} : = R_{2} - 3 R_{1}, R_{2} : = - \frac{1}{2} R_{2} and R_{1} : = R_{2} - 3 R_{2}

applied to the matrix in the question gives this matrix.

Not correct. Choice (d) is false.

Check your sequence of row operations.

Not correct. Choice (e) is false.

This matrix is not even is row echelon form!

Question 5

The matrix

[\begin{matrix} 0 & 1 & - 2 & 3 & 5 & - 7 \\ 0 & 0 & 1 & 0 & 1 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}]

is the augmented matrix of a system of linear equations. How many free parameters are needed for the solution of this system of equations?

Your answer is correct

The augmented matrix is already in echelon form, so the number of free parameters needed for the general solution is equal to the number of columns in the augmented matrix which do not contain a leading

1

for some row (in this case, columns 1, 4 and 5). You can also see this by looking at the corresponding system of equations. Let the variables be

x_{1}

x_{2}

x_{3}

x_{4}

and

x_{5}

then the correspding system of equations is:

\begin{array}{rcl} 0 x_{1} + 1 x_{2} - 2 x_{3} + 3 x_{4} + 5 x_{5} & = - 7 \\ 0 x_{1} + 0 x_{2} + 1 x_{3} + 0 x_{4} + 1 x_{5} & = 10 \\ 0 x_{1} + 0 x_{2} + 0 x_{3} + 0 x_{4} + 0 x_{5} & = 0 \end{array}

Consequently, there are 3 free parameters; namely, $x_{1}$ , $x_{4}$ and $x_{5}$ . Setting $x_{1} = p$ , $x_{4} = q$ , $x_{5} = r$ , we obtain the complete solution by back substitution:

\begin{array}{l} x_{3} & = 10 - x_{5} = 10 - r \\ x_{2} & = - 7 + 2 x_{3} - 3 x_{4} - 5 x_{5} = - 7 + 2 (10 - r) - 3 q - 5 r \\ = 13 - 3 q - 7 r \end{array}

That is, $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = (p, 13 - 3 q - 7 r, 10 - r, q, r)$ .

Not correct. You may try again.

Find the echelon form of the augmented matrix.

Question 6

Let

A

be the augmented matrix of the following system of equations.

\begin{array}{l} x + y + z & = 2, \\ x + 2 y + a z & = 4, \\ 2 x + 3 a y + 2 z & = 6, \end{array}

where

a

is some real number. Using elementary row operations, the matrix

A

can be reduced to the following matrix:

[\begin{matrix} 1 & 1 & 1 & 2 \\ 0 & 1 & a - 1 & 2 \\ 0 & 0 & (1 - a) (3 a - 2) & 6 (a - 1) \end{matrix}]

Which of the following statements are correct? (More than one answer may be correct.)

a)		There is exactly one value of $a$ for which the original system of equations has a unique solution.
b)		There are exactly two values of $a$ for which the original system of equations has a unique solution.
c)		There is exactly one value for $a$ for which the original system of equations does not have a unique solution.
d)		There are exactly two values for $a$ for which the original system of equations does not have a unique solution.
e)		There is exactly one value of $a$ for which the original system of equations is inconsistent.
f)		There are exactly two values of $a$ for which the original system of equations is inconsistent.

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

This system of equations has a unique solution for all values of

a

other than

1

2 ∕ 3

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

This system of equations has a unique solution for all values of

a

other than

1

2 ∕ 3

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

When

a = 1

the system has an infinite number of solutions and when

a = 2 ∕ 3

there are no solutions (the system is inconsistent).

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

When

a = 1

the system has an infinite number of solutions and when

a = 2 ∕ 3

there are no solutions (the system is inconsistent). For all other values of

a

the system has a unique solution.

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

a = 2 ∕ 3

then the reduced augmented matrix becomes

[\begin{matrix} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 2 \end{matrix}]

So the system of equations is inconsistent. For all other values of

a

the system is consistent.

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

The system is inconsistent only if

a = 2 ∕ 3

Your answers are correct

False. This system of equations has a unique solution for all values of $a$ other than $1$ or $2 ∕ 3$ .
False. This system of equations has a unique solution for all values of $a$ other than $1$ or $2 ∕ 3$ .
False. When $a = 1$ the system has an infinite number of solutions and when $a = 2 ∕ 3$ there are no solutions (the system is inconsistent).
True. When $a = 1$ the system has an infinite number of solutions and when $a = 2 ∕ 3$ there are no solutions (the system is inconsistent). For all other values of $a$ the system has a unique solution.
True. If $a = 2 ∕ 3$ then the reduced augmented matrix becomes $[\begin{matrix} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 2 \end{matrix}]$ So the system of equations is inconsistent. For all other values of $a$ the system is consistent.
False. The system is inconsistent only if $a = 2 ∕ 3$ .

Question 7

A matrix is a binary matrix if all of its entries are

0

1

. For example, these two matrices

[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] and [\begin{matrix} 0 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}]

are binary matrices, wheres the matrix

[\begin{matrix} 0 & - 1 & 1 \\ 0 & - 1 & 1 \end{matrix}]

is not binary. How many binary 2 by 3 row echelon matrices are there ?

Your answer is correct

The 22 binary row echelon matrices are:

\begin{matrix} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}] \\ [\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix}] \\ [\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] & [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}] \\ [\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] . \end{matrix}

Not correct. You may try again.

You can list the row echelon binary matrices systematically as they all have one of the following forms:

\begin{matrix} [\begin{matrix} 1 & * & * \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & * & * \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 1 & * & * \\ 0 & 1 & * \end{matrix}] & [\begin{matrix} 0 & 1 & * \\ 0 & 0 & 0 \end{matrix}] \\ [\begin{matrix} 0 & 1 & * \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \end{matrix}

where the *’s are either $0$ or $1$ .

Question 8

As in question 7, call a matrix binary if all of its entires are

0

1

. (See question 7 for some examples.) How many binary 2 by 3 reduced row echelon matrices are there ?

Your answer is correct

The 15 reduced echelon binary matrices are

\begin{matrix} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}] \\ [\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix}] \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] . \end{matrix}

Not correct. You may try again.

You can list the row echelon binary matrices systematically as they all have one of the following forms:

\begin{matrix} [\begin{matrix} 1 & * & * \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & * & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 1 & 0 & * \\ 0 & 1 & * \end{matrix}] & [\begin{matrix} 0 & 1 & * \\ 0 & 0 & 0 \end{matrix}] \\ [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] & [\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \end{matrix}

where the *’s are either $0$ or $1$ .

Question 9

The (unbalanced) following chemical equation

{C u}_{2} S + O_{2} + C \to {S O}_{2} + C u + C O .

describes how copper (Cu) can be extracted from chalcocite (

{Cu}_{2} S

) by combining it with carbon (C) and oxygen (

O_{2}

) to produce copper (Cu), sulfur dioxid (

{SO}_{2}

) and carbon monoxide (CO).
The chemical equation above is unbalanced because, for example, there are two copper atoms on the left hand side and only one copper atom on the right hand side. Which of the following augmented matrices corresponds to the system of linear equations that you would use in order to balance the equation above?

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

Not correct. Choice (a) is false.

We you to find integers

n_{1}

n_{2}, \dots, n_{6}

such that

n_{1} {C u}_{2} S + n_{2} O_{2} + n_{3} C \to n_{4} {S O}_{2} + n_{5} C u + n_{6} C O .

To balance the equation we need the same number of copper atoms on both sides so

2 n_{1} = n_{5}

, etc.

Your answer is correct.

We need to find integers

n_{1}

n_{2}, \dots, n_{6}

such that

n_{1} {C u}_{2} S + n_{2} O_{2} + n_{3} C \to n_{4} {S O}_{2} + n_{5} C u + n_{6} C O

In order to balance each element we need to have the same number of atoms of each type on both the left and right hand side of the equation. That is, the following equations must hold:

\begin{matrix} Element & Balancing constraint \\ C u & 2 n_{1} & = & n_{5} \\ S & n_{1} & = & n_{4} \\ O & 2 n_{2} & = & 2 n_{4} + n_{6} \\ C & n_{3} & = & n_{6} \end{matrix}

For the augmented matrix we need the variables on the left and the constants on the right. That is, we have the equations:

\begin{array}{l} 2 n_{1} - n_{5} & = 0 \\ n_{1} - n_{4} & = 0 \\ 2 n_{2} - 2 n_{4} - n_{6} & = 0 \\ n_{3} - n_{6} & = 0 \end{array}

Using the augmented matrix easy to see that one way to balance this equation is

{C u}_{2} S + 2 O_{2} + 2 C \to {S O}_{2} + 2 C u + 2 C O

Not correct. Choice (c) is false.

We you to find integers

n_{1}

n_{2}, \dots, n_{6}

such that

n_{1} {C u}_{2} S + n_{2} O_{2} + n_{3} C \to n_{4} {S O}_{2} + n_{5} C u + n_{6} C O .

To balance the equation we need the same number of copper atoms on both sides so

2 n_{1} = n_{5}

, etc.

Not correct. Choice (d) is false.

We you to find integers

n_{1}

n_{2}, \dots, n_{6}

such that

n_{1} {C u}_{2} S + n_{2} O_{2} + n_{3} C \to n_{4} {S O}_{2} + n_{5} C u + n_{6} C O .

To balance the equation we need the same number of copper atoms on both sides so

2 n_{1} = n_{5}

, etc.

Question 10

Find the augmented matrices associated with finding the currents

I_{1}

I_{2}

and

I_{3}

in the circuit below.

Hints :

The total current flowing into a node equals the total current flowing out.
In each loop the sum of the voltage drops across the circuit elements is zero.
The voltage drop across a resistor with resistance $R$ and with a current $I$ passing through it is $V = I R$ .

Not correct. Choice (a) is false.

Your answer is correct.

The current into node

A

I_{3}

and the current out is

I_{1} + I_{2}

I_{1} + I_{2} - I_{3} = 0 .

In the top loop the voltage drop is zero so

I_{3} + 40 I_{3} + 30 I_{1} - 45 = 0

implying

30 I_{1} + 41 I_{3} = 45 .

In the bottom loop the voltage drop is also zero implying

- 80 + I_{2} + 20 I_{2} - 30 I_{1} = 0

(be careful with the direction of the current). The voltage drop across the 30

Ω

resistor is negative since the current is labelled in the opposite direction thus

30 I_{1} - 21 I_{2} - 80 = 0 .

Hence an augmented matrix is

[\begin{matrix} 1 & 1 & - 1 & 0 \\ 30 & 0 & 41 & 45 \\ 30 & - 21 & 0 & 80 \end{matrix}]

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Questions:

right first attempt
right
wrong

The University of Sydney - School of Mathematics and Statistics

Quiz 6: Reduced row echelon form

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Sitemap

a)		$[\begin{matrix} 1 & 4 & 0 & 5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$		b)		$[\begin{matrix} 1 & 4 & - 7 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$
c)		$[\begin{matrix} 1 & 4 & 0 & 5 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$		d)		$[\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$

a)		$R_{1} : = R_{1} - R_{3}$ , $R_{2} : = R_{2} - R_{1}$ , $R_{3} : = R_{3} - R_{1}$ and $R_{2} : = \frac{1}{4} R_{2}$ .
b)		$R_{1} \leftrightarrow R_{2}$ , $R_{2} : = R_{2} - 2 R_{1}$ , $R_{3} : = R_{3} - R_{1}$ , $R_{2} : = - R_{2}$ and $R_{3} = R_{3} - 3 R_{2}$ .
c)		$R_{1} : = \frac{1}{2} R_{1}$ , $R_{2} : = R_{2} - R_{1}$ , $R_{3} = R_{3} - R_{1}$ , $R_{3} : = R_{3} - 7 R_{2}$ and $R_{2} : = 2 R_{2}$ .

a)	$[\begin{matrix} 1 & 1 & - 1 & 0 \\ 30 & 0 & 41 & 45 \\ 30 & 21 & 0 & 80 \end{matrix}]$	b)	$[\begin{matrix} 1 & 1 & - 1 & 0 \\ 30 & 0 & 41 & 45 \\ 30 & - 21 & 0 & 80 \end{matrix}]$
c)	$[\begin{matrix} 1 & - 1 & 1 & 0 \\ 30 & 0 & 41 & - 45 \\ 30 & 21 & 0 & 80 \end{matrix}]$	d)	$[\begin{matrix} 1 & - 1 & - 1 & 0 \\ 30 & 0 & 21 & 45 \\ 0 & 21 & 41 & 35 \end{matrix}]$