Select all the matrices which are in row echelon form. (Zero or more options can be
correct)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

For example, choice (e) should be True.

*There is at least one mistake.*

For example, choice (a) should be False.

This is not in row echelon form because the leading terms in rows 1 and 3 are not equal
to 1.

*There is at least one mistake.*

For example, choice (b) should be False.

This is not in row echelon form because the leading term in the second
row is not further to the right than the leading term in the first row.

*There is at least one mistake.*

For example, choice (c) should be False.

This
is not in row echelon form because the row of zeros is not underneath all the non-zero
rows.

*There is at least one mistake.*

For example, choice (d) should be True.

There is nothing in the definition of row echelon form that says the matrix must have
some non-zero entries. The zero matrix of any size is always in row echelon form.

*There is at least one mistake.*

For example, choice (e) should be True.

*Correct!*

*False*This is not in row echelon form because the leading terms in rows 1 and 3 are not equal to 1.*False*This is not in row echelon form because the leading term in the second row is not further to the right than the leading term in the first row.*False*This is not in row echelon form because the row of zeros is not underneath all the non-zero rows.*True*There is nothing in the definition of row echelon form that says the matrix must have some non-zero entries. The zero matrix of any size is always in row echelon form.*True*

Select all the matrices which are in reduced row echelon form. (Zero or more options can
be correct)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be True.

For example, choice (d) should be True.

*There is at least one mistake.*

For example, choice (a) should be False.

The leading 1 in row 3 has a non-zero entry above it in the
same column, so this matrix is not in reduced row echelon form.

*There is at least one mistake.*

For example, choice (b) should be False.

This
is not in reduced row echelon form because the row of zeros is not underneath all the
non-zero rows.

*There is at least one mistake.*

For example, choice (c) should be True.

*There is at least one mistake.*

For example, choice (d) should be True.

*Correct!*

*False*The leading 1 in row 3 has a non-zero entry above it in the same column, so this matrix is not in reduced row echelon form.*False*This is not in reduced row echelon form because the row of zeros is not underneath all the non-zero rows.*True**True*

One more row operation must be applied to
$\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$
in order to obtain a matrix in row echelon form. Which operation
is it? Exactly one option must be correct)

*Choice (a) is incorrect*

Do the row
operation and you’ll see why the resulting matrix is not in row echelon form!

*Choice (b) is incorrect*

Do the row operation and you’ll see why the resulting matrix is not in row echelon
form!

*Choice (c) is incorrect*

Do the row operation and you’ll see why the resulting matrix is not in row echelon
form!

*Choice (d) is correct!*

In trying to find out if the lines (in 3D space)
$$\mathbf{x}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \\ \hfill 4\hfill \end{array}\right]+s\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\mathbf{x}=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 5\hfill \\ \hfill 1\hfill \end{array}\right]+t\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill 3\hfill \end{array}\right]$$
have a common point of intersection, we solve a system of 3 equations in the two
unknowns $s$
and $t$. After
applying a number of elementary row operations (starting with the augmented matrix) we
obtain $\left[\begin{array}{c}\hfill \begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right].$
What conclusion can be drawn about the point of intersection?
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

*Choice (e) is correct!*

The third row of the matrix corresponds to the equation
$0s+0t=1$, that is,
that $0=1$.
This shows that the system is inconsistent. There are no solutions for
$s$ and
$t$ and
hence the two lines do not meet.

*Choice (f) is incorrect*

Consider the matrix $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right].$
Which is the most sensible elementary row operation to perform next, if we want to progress
towards a matrix in row echelon form? Exactly one option must be correct)

*Choice (a) is correct!*

In working towards row echelon form, we move through the matrix from left to right
and from top to bottom. Once the leading entry is in place in row 1, we
make all entries underneath it equal to zero. Therefore we should clear the
entry in row 3, column 1 before we do anything else.

*Choice (b) is incorrect*

This
will make a zero in the second entry of row 3, but this is not necessary for row
echelon form.

*Choice (c) is incorrect*

This will not give us an efficient path
to row echelon form because the position currently occupied by 3 in the
first column will still need to be cleared to zero.

*Choice (d) is incorrect*

Consider the matrix $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right].$
Which is the most sensible elementary row operation to perform
next, if we want to progress towards a matrix in reduced row echelon
form? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

This will produce the matrix
$\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right].$
Reduced row echelon form can be obtained with just two more elementary row
operation. (Can you see what they are?)

For which value of $k$
is the system
$$\begin{array}{llll}\hfill x+2y& =k\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-ky& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
inconsistent (that is, has no solution)? Exactly one option must be correct)

*Choice (a) is incorrect*

Hint: write down the augmented matrix of the system and apply a row
operation to work towards row echelon form. Then search for the value of
$k$ that makes the
system inconsistent.

*Choice (b) is correct!*

The augmented matrix is row equivalent to the matrix
$\left[\begin{array}{c}\hfill \begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill k\hfill \\ \hfill 0\hfill & \hfill -k-4\hfill & \hfill 4-2k\hfill \end{array}\hfill \end{array}\right].$Therefore if
$k=-4$, the last row becomes
$\left[0\phantom{\rule{1em}{0ex}}0\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}12\right]$, indicating an
inconsistent system.

*Choice (c) is incorrect*

Hint: write down the augmented matrix of the system and apply a row
operation to work towards row echelon form. Then search for the value of
$k$ that makes the
system inconsistent.

*Choice (d) is incorrect*

Hint: write down the augmented matrix of the system and apply a row
operation to work towards row echelon form. Then search for the value of
$k$ that
makes the system inconsistent.

For which value of $a$
does the system
$$\begin{array}{llll}\hfill x+ay+z& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ax+y+2z& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-2y+z& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
have infinitely many solutions? Exactly one option must be correct)

*Choice (a) is incorrect*

Hint: write down the augmented matrix of the system and apply a number
of row operations (working towards row echelon form) until the matrix is
simplified. (The matrix will continue to contain various terms involving
$a$.) Then search
for the value of $a$
that gives just two non-zero rows in the matrix.

*Choice (b) is incorrect*

Hint: write down the augmented matrix of the system and apply a number
of row operations (working towards row echelon form) until the matrix is
simplified. (The matrix will continue to contain various terms involving
$a$.) Then search
for the value of $a$
that gives just two non-zero rows in the matrix.

*Choice (c) is incorrect*

Hint: write down the augmented matrix of the system and apply a number
of row operations (working towards row echelon form) until the matrix is
simplified. (The matrix will continue to contain various terms involving
$a$.) Then search
for the value of $a$
that gives just two non-zero rows in the matrix.

*Choice (d) is correct!*

The
augmented matrix of the system can be reduced to
$$\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill a\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1-{a}^{2}\hfill & \hfill 2-a\hfill & \hfill -a\hfill \\ \hfill 0\hfill & \hfill -2-a\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\hfill \end{array}\right].$$
Therefore when $a=-2$,
the last row becomes a row of zeros, and one parameter is needed to describe the
solutions. That is, the system has infinitely many solutions.

For which values of $k$
will the system
$$\begin{array}{llll}\hfill x+y+z& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-3y+z& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill kx+2y+kz& =3k\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
have a unique solution? Exactly one option must be correct)

*Choice (a) is correct!*

The
augmented matrix
$$\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill -3\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill k\hfill & \hfill 2\hfill & \hfill k\hfill & \hfill 3k\hfill \end{array}\hfill \end{array}\right]$$
reduces (after the application of a number of row operations) to
$$\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \frac{1}{5}\hfill & \hfill \frac{9}{5}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{k-2}{5}\hfill & \hfill -\frac{k+18}{5}\hfill \end{array}\hfill \end{array}\right].$$
The leading entry in row 3 is a non-zero number if and only if
$k\ne 2$. Assuming
that $k\ne 2$, we
have three non-zero leading entries in rows 1, 2 and 3, leading to a unique solution.

*Choice (b) is incorrect*

Hint: write down the augmented matrix of the system and apply a number of row
operations, working towards row echelon form. Then search for values of
$k$ which
guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column 3)
positions.

*Choice (c) is incorrect*

Hint: write down the augmented matrix of the system and apply a number of row
operations, working towards row echelon form. Then search for values of
$k$ which
guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column 3)
positions.

*Choice (d) is incorrect*

Hint: write down the augmented matrix of the system and apply a number of row
operations, working towards row echelon form. Then search for values of
$k$ which
guarantee leading terms in (row 1, column 1), (row 2, column 2) and (row 3, column
3) positions.

Select all the true statements from the options below. (Zero or more options can be
correct)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

For example, choice (e) should be False.

For example, choice (f) should be True.

For example, choice (g) should be False.

For example, choice (h) should be True.

*There is at least one mistake.*

For example, choice (a) should be False.

The system could be inconsistent (that
is, have no solution).

*There is at least one mistake.*

For example, choice (b) should be False.

The system
$$\begin{array}{llll}\hfill x+y+z& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x+2y+2z& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
has 2 equations and 3 unknowns but is clearly inconsistent.

*There is at least one mistake.*

For example, choice (c) should be False.

The system
$$\begin{array}{llll}\hfill x+y& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x+2y& =8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 5x+5y& =20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
has more equations than unknowns but has infinitely many solutions.

*There is at least one mistake.*

For example, choice (d) should be True.

*There is at least one mistake.*

For example, choice (e) should be False.

The matrix
$\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ is in
row echelon form but not reduced row echelon form.

*There is at least one mistake.*

For example, choice (f) should be True.

*There is at least one mistake.*

For example, choice (g) should be False.

In fact, there are
fifteen such matrices. You can find them systematically by first counting how many
have two leading 1’s (there are seven of these), then how many have one
leading 1 (there are also seven of these) and then no leading 1 (there is one of
these, the zero matrix). Of the seven with two leading 1s, four have the form
$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill b\hfill \end{array}\right]$ where
$a$ and
$b$ can be
either $0$ or
$1$, two have
the form $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$
where $a$ can
be either $0$
or $1$, and
the last is $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$.
You can verify in a similar way that there are seven with one leading 1.

*There is at least one mistake.*

For example, choice (h) should be True.

*Correct!*

*False*The system could be inconsistent (that is, have no solution).*False*The system $$\begin{array}{llll}\hfill x+y+z& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x+2y+2z& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ has 2 equations and 3 unknowns but is clearly inconsistent.*False*The system $$\begin{array}{llll}\hfill x+y& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x+2y& =8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 5x+5y& =20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ has more equations than unknowns but has infinitely many solutions.*True**False*The matrix $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ is in row echelon form but not reduced row echelon form.*True**False*In fact, there are fifteen such matrices. You can find them systematically by first counting how many have two leading 1’s (there are seven of these), then how many have one leading 1 (there are also seven of these) and then no leading 1 (there is one of these, the zero matrix). Of the seven with two leading 1s, four have the form $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill b\hfill \end{array}\right]$ where $a$ and $b$ can be either $0$ or $1$, two have the form $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ where $a$ can be either $0$ or $1$, and the last is $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$. You can verify in a similar way that there are seven with one leading 1.*True*