Which of the following equations is/are linear in the variables
$x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$?
(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.

For example, choice (e) should be False.

For example, choice (f) should be True.

*There is at least one mistake.*

For example, choice (a) should be False.

In a linear equation the variables cannot occur in the form of squares. When
$x$ occurs it must
be in the form $kx$ for
some constant $k$
(and the other terms must also be in this form).

*There is at least one mistake.*

For example, choice (b) should be False.

Terms
such as ${e}^{y}$
are not permitted - the general form of a linear equation in
$x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ is
$ax+by+cz=d$ for some
constants $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.

*There is at least one mistake.*

For example, choice (c) should be True.

This equation is
linear in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ because it
is in the form $ax+by+cz=d$ for
some constants $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.

*There is at least one mistake.*

For example, choice (d) should be True.

This is a linear
equation in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ since
we can think of it as $x+0y+78z=31.$

*There is at least one mistake.*

For example, choice (e) should be False.

This equation is
linear in the variables $u=sinx$,
$v=cosy$ and
$w=tanz$ but is not
linear in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$.

*There is at least one mistake.*

For example, choice (f) should be True.

This is a linear
equation in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ since
we can think of it as $x-2y+0z=0.$

*Correct!*

*False*In a linear equation the variables cannot occur in the form of squares. When $x$ occurs it must be in the form $kx$ for some constant $k$ (and the other terms must also be in this form).*False*Terms such as ${e}^{y}$ are not permitted - the general form of a linear equation in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ is $ax+by+cz=d$ for some constants $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.*True*This equation is linear in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ because it is in the form $ax+by+cz=d$ for some constants $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.*True*This is a linear equation in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ since we can think of it as $x+0y+78z=31.$*False*This equation is linear in the variables $u=sinx$, $v=cosy$ and $w=tanz$ but is not linear in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$.*True*This is a linear equation in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$ since we can think of it as $x-2y+0z=0.$

Which matrix is the augmented matrix of the linear system
$$\begin{array}{llll}\hfill 3x+y+z-t& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+y& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y-2t& =7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
in the variables $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z,\phantom{\rule{0.3em}{0ex}}t$? Exactly
one option must be correct)

*Choice (a) is correct!*

*Choice (b) is incorrect*

The second row corresponds to the equation
$y+t=1$, which is not the correct
second equation.

*Choice (c) is incorrect*

The third row corresponds to the equation
$x-2y=7$, which is not the correct third
equation.

*Choice (d) is incorrect*

The third row
corresponds to the equation $y-2z=7$,
which is not the correct third equation.

Which system of equations corresponds to the following augmented matrix?
$$\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\hfill \end{array}\right]$$
Exactly one option must be correct)

*Choice (a) is incorrect*

Check the last row of the augmented matrix, in particular the signs!

*Choice (b) is incorrect*

Check the second and third rows of the augmented matrix.

*Choice (c) is incorrect*

Check the first row of the augmented matrix.

*Choice (d) is correct!*

Which two linear systems have exactly the same (unique) solution?
(Zero or more options can be correct)

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be True.

*There is at least one mistake.*

For example, choice (a) should be True.

*There is at least one mistake.*

For example, choice (b) should be False.

The solution to this system is $x=2,\phantom{\rule{0.3em}{0ex}}y=1$
which is not a solution to either of the other two systems.

*There is at least one mistake.*

For example, choice (c) should be True.

*Correct!*

*True**False*The solution to this system is $x=2,\phantom{\rule{0.3em}{0ex}}y=1$ which is not a solution to either of the other two systems.*True*

Find the solution of the system
$$\begin{array}{llll}\hfill x+5y+7z& =-13\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2y-z& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3z& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
by back substitution. 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!*

Which row operation should be performed next, in order to efficiently solve (by back
substitution) the system of equations corresponding to the following augmented
matrix?
$$\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right]$$
Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

This is the best move because it will produce a new row 3
with entries $\left[0\phantom{\rule{1em}{0ex}}0\phantom{\rule{1em}{0ex}}-5\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}-7\right]$
from which we immediately see that the third variable is equal to
$7\u22155$. The
remaining variables can now be found by back substitution.

*Choice (d) is incorrect*

Which option is the complete set of solutions of the equation
$x+2y-4z=13$, written in
parametric form with $s$
and $t$
as the parameters? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

Let
$x=s$ and
$y=t$. Then
since $z=\frac{1}{4}\left(x+2y-13\right)$, we
must have $z=\frac{1}{4}\left(s+2t-13\right)=-\frac{13}{4}+\frac{s}{4}+\frac{t}{2}$.

There is a unique solution of the following linear system. What is it?
$$\begin{array}{llll}\hfill {x}_{1}+{x}_{2}-3{x}_{3}& =-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -7{x}_{2}+7{x}_{3}& =7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3{x}_{3}& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
Exactly one option must be correct)

*Choice (a) is correct!*

These values satisfy all three equations.

*Choice (b) is incorrect*

The first equation
is not satisfied.

*Choice (c) is incorrect*

The first and second equations are not satisfied.

*Choice (d) is incorrect*

None
of the equations are satisfied.

The following augmented matrices represent different linear systems in
$x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$.
Which of the linear systems has a unique solution?
$$A.\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 3\hfill & \hfill 30\hfill & \hfill 90\hfill & \hfill 120\hfill \\ \hfill 1\hfill & \hfill 10\hfill & \hfill 30\hfill & \hfill 40\hfill \\ \hfill 2\hfill & \hfill 20\hfill & \hfill 60\hfill & \hfill 80\hfill \end{array}\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}B.\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 12\hfill & \hfill 1\hfill & \hfill 5\hfill & \hfill 15\hfill \\ \hfill 11\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 14\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}C.\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 5\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right]$$
Exactly one option must be correct)

*Choice (a) is incorrect*

System
$A$ must have
infinitely many solutions. Each row of the augmented matrix is a multiple of the same row,
namely $\left[1\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}10\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}30\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}40\right]$.
So we have essentially just one equation linking the three variable in system
$A$. This
means that two parameters are needed to specify a complete set of solutions; that is
$A$ does not have a
unique solution.

*Choice (b) is incorrect*

This can’t be the right answer because in system
$B$, the first
row is the sum of the second and third rows. This means that the row operation on
${R}_{1}$ which
subtracts ${R}_{2}$ from
${R}_{1}$ will produce a new
first row identical to ${R}_{3}$.
This means that instead of having three independent equations linking
the three unknown variables, we have only two. In fact, system
$B$ has infinitely many
solutions.

*Choice (c) is correct!*

The
unique solution is $x=-4,\phantom{\rule{0.3em}{0ex}}y=1,\phantom{\rule{0.3em}{0ex}}z=1$.

*Choice (d) is incorrect*

Two of the following augmented matrices correspond to linear systems with no solution
(inconsistent systems). Which two are they? (Zero or more options can be correct)

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be True.

*There is at least one mistake.*

For example, choice (a) should be True.

This represents an inconsistent system. The second row corresponds to the equation
$3x+y=1$. However the third
row corresponds to $3x+y=2$.
Clearly these are contradictory statements and therefore the system has no
solution.

*There is at least one mistake.*

For example, choice (b) should be False.

This system has the
unique solution $z=2,\phantom{\rule{0.3em}{0ex}}y=8,\phantom{\rule{0.3em}{0ex}}x=-3$, found
by back substitution.

*There is at least one mistake.*

For example, choice (c) should be False.

This system has infinitely many solutions, as the first row is the sum of
the second and third rows. There are only two independent equations
involving the three unknowns, leading to infinitely many solutions.

*There is at least one mistake.*

For example, choice (d) should be True.

This
represents an inconsistent system. If we multiply the first row by 4 we obtain
$\left[4\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}12\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}8\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}8\right]$ which corresponds to the
equation $4x+12y+8z=8$. However the
second row corresponds to $4x+12y+8z=7$.
Clearly these are contradictory statements and therefore the system has no solution.

*Correct!*

*True*This represents an inconsistent system. The second row corresponds to the equation $3x+y=1$. However the third row corresponds to $3x+y=2$. Clearly these are contradictory statements and therefore the system has no solution.*False*This system has the unique solution $z=2,\phantom{\rule{0.3em}{0ex}}y=8,\phantom{\rule{0.3em}{0ex}}x=-3$, found by back substitution.*False*This system has infinitely many solutions, as the first row is the sum of the second and third rows. There are only two independent equations involving the three unknowns, leading to infinitely many solutions.*True*This represents an inconsistent system. If we multiply the first row by 4 we obtain $\left[4\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}12\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}8\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}8\right]$ which corresponds to the equation $4x+12y+8z=8$. However the second row corresponds to $4x+12y+8z=7$. Clearly these are contradictory statements and therefore the system has no solution.