## MATH1014 Quizzes

Quiz 5: Systems of Linear Equations
Question 1 Questions
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)
 a) ${x}^{2}+2y+7z=3$ b) $\sqrt{2}\phantom{\rule{0.3em}{0ex}}x+{e}^{y}+z=0$ c) $3x-y-z=0$ d) $x+78z=31$ e) $sinx+3cosy-tanz=0$ f) $x=2y$

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!
1. 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).
2. 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 $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.
3. 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 $a,\phantom{\rule{0.3em}{0ex}}b,\phantom{\rule{0.3em}{0ex}}c,\phantom{\rule{0.3em}{0ex}}d$.
4. 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 $x+0y+78z=31.$
5. False This equation is linear in the variables $u=sinx$, $v=cosy$ and $w=tanz$ but is not linear in $x,\phantom{\rule{0.3em}{0ex}}y,\phantom{\rule{0.3em}{0ex}}z$.
6. 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)
 a) $\left[\begin{array}{c}\hfill \begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\hfill \end{array}\right]$ b) $\left[\begin{array}{c}\hfill \begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\hfill \end{array}\right]$ c) $\left[\begin{array}{c}\hfill \begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 7\hfill \end{array}\hfill \end{array}\right]$ d) $\left[\begin{array}{c}\hfill \begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 7\hfill \end{array}\hfill \end{array}\right]$

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)
 a) $\begin{array}{llll}\hfill a+2b& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill b+3c& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill b+c& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ b) $\begin{array}{llll}\hfill x+2z& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+3y& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -x-y& =-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ c) $\begin{array}{llll}\hfill u+2v& =3w+3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v+3w& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill v+w& =-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ d) $\begin{array}{llll}\hfill r+2s& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill s+3t& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -s-t& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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)
 a) $\begin{array}{llll}\hfill x+2y& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x-3y& =-4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ b) $\begin{array}{llll}\hfill 3x-4y& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-3y& =-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ c) $\begin{array}{llll}\hfill 5x+y& =7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 6x-2y& =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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!
1. True
2. 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.
3. 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)
 a) $x=13,\phantom{\rule{0.3em}{0ex}}y=1,\phantom{\rule{0.3em}{0ex}}z=3.$ b) $x=8,\phantom{\rule{0.3em}{0ex}}y=-1,\phantom{\rule{0.3em}{0ex}}z=3.$ c) $x=23,\phantom{\rule{0.3em}{0ex}}y=1,\phantom{\rule{0.3em}{0ex}}z=-3.$ d) $x=3,\phantom{\rule{0.3em}{0ex}}y=-1,\phantom{\rule{0.3em}{0ex}}z=3.$ e) $x=-39,\phantom{\rule{0.3em}{0ex}}y=1,\phantom{\rule{0.3em}{0ex}}z=3.$

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)
 a) An operation on row 1: ${R}_{1}-{R}_{2}$, that is, subtract row 2 from row 1. b) An operation on row 2: ${R}_{2}-\frac{1}{2}{R}_{3}$, that is, subtract half of row 3 from row 2. c) An operation on row 3: ${R}_{3}-2{R}_{2}$, that is, subtract twice row 2 from row 3. d) An operation on row 3: ${R}_{3}-2{R}_{1}$, that is, subtract twice row 1 from row 3.

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 $7∕5$. 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)
 a) $\left[s,\phantom{\rule{0.3em}{0ex}}-2s+13,\phantom{\rule{0.3em}{0ex}}4t+13\right]$ b) $\left[t,\phantom{\rule{0.3em}{0ex}}s,\phantom{\rule{0.3em}{0ex}}-\frac{13}{4}+\frac{s}{4}+\frac{t}{2}\right]$ c) $\left[s,\phantom{\rule{0.3em}{0ex}}t,\phantom{\rule{0.3em}{0ex}}\frac{13}{4}-\frac{s}{4}-\frac{t}{2}\right]$ d) $\left[s,\phantom{\rule{0.3em}{0ex}}t,\phantom{\rule{0.3em}{0ex}}-\frac{13}{4}+\frac{s}{4}+\frac{t}{2}\right]$

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)
 a) ${x}_{1}=1,\phantom{\rule{0.3em}{0ex}}{x}_{2}=2,\phantom{\rule{0.3em}{0ex}}{x}_{3}=3.$ b) ${x}_{1}=-1,\phantom{\rule{0.3em}{0ex}}{x}_{2}=2,\phantom{\rule{0.3em}{0ex}}{x}_{3}=3.$ c) ${x}_{1}=-1,\phantom{\rule{0.3em}{0ex}}{x}_{2}=-2,\phantom{\rule{0.3em}{0ex}}{x}_{3}=3.$ d) ${x}_{1}=1,\phantom{\rule{0.3em}{0ex}}{x}_{2}=2,\phantom{\rule{0.3em}{0ex}}{x}_{3}=-3.$

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)
 a) $A$ only. b) $B$ only. c) $C$ only. d) None has a unique solution.

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)
 a) $\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\hfill \end{array}\right]$ b) $\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill & \hfill 15\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 14\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 4\hfill \end{array}\hfill \end{array}\right]$ c) $\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right]$ d) $\left[\begin{array}{c}\hfill \begin{array}{cccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 12\hfill & \hfill 8\hfill & \hfill 7\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\hfill \end{array}\right]$

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!
1. 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.
2. 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.
3. 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.
4. 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.