## MATH1014 Quizzes

Quiz 6: Solving Linear Systems
Question 1 Questions
Select all the matrices which are in row echelon form. (Zero or more options can be correct)
 a) $\left[\begin{array}{ccccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right]$ b) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ c) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ d) $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ e) $\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

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!
1. False This is not in row echelon form because the leading terms in rows 1 and 3 are not equal to 1.
2. 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.
3. False This is not in row echelon form because the row of zeros is not underneath all the non-zero rows.
4. 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.
5. True
Select all the matrices which are in reduced row echelon form. (Zero or more options can be correct)
 a) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 7\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ b) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ c) $\left[\begin{array}{cccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ d) $\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 11\hfill \end{array}\right]$

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!
1. 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.
2. False This is not in reduced row echelon form because the row of zeros is not underneath all the non-zero rows.
3. True
4. 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)
 a) An operation on row 2: ${R}_{2}-{R}_{3}$ b) An operation on row 3: ${R}_{3}-{R}_{1}$ c) An operation on row 3: ${R}_{3}-{R}_{2}$ d) Swap rows 2 and 3: ${R}_{2}↔{R}_{3}$.

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)
 a) The point of intersection is $\left(-1,-3,1\right)$. b) The point of intersection is $\left(1,3,-1\right)$. c) The point of intersection is $\left(-1,-3,0\right)$. d) The point of intersection is $\left(1,3,0\right)$. e) The two lines do not intersect. f) There is not enough information given to answer the question.

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)
 a) An operation on row 3: ${R}_{3}-3{R}_{1}$. b) An operation on row 3: ${R}_{3}-\frac{1}{2}{R}_{2}$. c) Swap rows 2 and 3. d) An operation on row 2: ${R}_{2}-2{R}_{3}$.

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)
 a) An operation on row 2: $\frac{1}{3}{R}_{2}$. b) An operation on row 2: ${R}_{2}-3{R}_{3}$. c) An operation on row 1: ${R}_{1}-{R}_{2}$. d) Swap rows 2 and 3.

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)
 a) $k=0$ b) $k=-4$ c) $k=2$ d) $k=-1$

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)
 a) $a=1$ b) $a=-3$ c) $a=5$ d) $a=-2$

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)
 a) All $k$ except $k=2$. b) All values of $k$ c) $k=2$ d) $k=-2$ and $2$.

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)
 a) Every linear system has at least one solution. b) A linear system with 2 equations in 3 unknowns always has infinitely many solutions. c) A linear system with more equations than unknowns is always inconsistent. d) The reduced row echelon form of any matrix is unique. e) Every row echelon matrix is also a reduced row echelon matrix. f) Every linear system has either 0, 1 or infinitely many solutions. g) There are five different reduced row echelon matrices with 2 rows and 3 columns, in which each entry is either 0 or 1. h) A matrix may have more than one row echelon form.

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!
1. False The system could be inconsistent (that is, have no solution).
2. 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.
3. 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.
4. True
5. 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.
6. True
7. 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.
8. True