## MATH1002 Quizzes

Quiz 8: Inverses and elementary matrices
Question 1 Questions
Let $A=\left[\begin{array}{cc}\hfill -5\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -4\hfill \end{array}\right]$. Which of the following statements are true: (Zero or more options can be correct)
 a) $A$ is singular b) $A$ is invertible c) $A$ is non-singular d) ${A}^{-1}$ does not exist e) The inverse of $A$ is $\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]$

There is at least one mistake.
For example, choice (a) should be False.
$A$ is singular if ${A}^{-1}$ does not exist
There is at least one mistake.
For example, choice (b) should be True.
$A$ is invertible is the same as saying that $A$ is non-singular, or that ${A}^{-1}$ exists. We can see that $A$ is invertible by applying elementary row operations: $\left[A\mid I\right]=\left[\begin{array}{cccc}\hfill -5& \hfill 7& \hfill 1& \hfill 0\\ \hfill 3& \hfill -4& \hfill 0& \hfill 1\end{array}\right]\to \left[\begin{array}{cccc}\hfill 1& \hfill -1& \hfill 1& \hfill 2\\ \hfill 3& \hfill -4& \hfill 0& \hfill 1\end{array}\right]\to \left[\begin{array}{cccc}\hfill 1& \hfill -1& \hfill 1& \hfill 2\\ \hfill 0& \hfill -1& \hfill -3& \hfill -5\end{array}\right].$
There is at least one mistake.
For example, choice (c) should be True.
$A$ is non-singular is the same as saying that $A$ is invertible, or that ${A}^{-1}$ exists.
There is at least one mistake.
For example, choice (d) should be False.
You can see that $A$ is invertible using elementary row operations.
There is at least one mistake.
For example, choice (e) should be True.
You can check that this matrix is the inverse of $A$ by showing that $A\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]={I}_{2}=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]A$.
Correct!
1. False $A$ is singular if ${A}^{-1}$ does not exist
2. True $A$ is invertible is the same as saying that $A$ is non-singular, or that ${A}^{-1}$ exists. We can see that $A$ is invertible by applying elementary row operations: $\left[A\mid I\right]=\left[\begin{array}{cccc}\hfill -5& \hfill 7& \hfill 1& \hfill 0\\ \hfill 3& \hfill -4& \hfill 0& \hfill 1\end{array}\right]\to \left[\begin{array}{cccc}\hfill 1& \hfill -1& \hfill 1& \hfill 2\\ \hfill 3& \hfill -4& \hfill 0& \hfill 1\end{array}\right]\to \left[\begin{array}{cccc}\hfill 1& \hfill -1& \hfill 1& \hfill 2\\ \hfill 0& \hfill -1& \hfill -3& \hfill -5\end{array}\right].$
3. True $A$ is non-singular is the same as saying that $A$ is invertible, or that ${A}^{-1}$ exists.
4. False You can see that $A$ is invertible using elementary row operations.
5. True You can check that this matrix is the inverse of $A$ by showing that $A\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]={I}_{2}=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]A$.
Let $A=\left[\begin{array}{ccc}\hfill -3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right]$. The inverse of $A$ (i.e., ${A}^{-1}$) is: Exactly one option must be correct)
 a) undefined b) $B=\left[\begin{array}{ccc}\hfill -3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right]$ c) $C=\left[\begin{array}{ccc}\hfill -\frac{1}{3}\hfill & \hfill 1\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill \hfill \\ \hfill \frac{1}{2}\hfill & \hfill \frac{1}{3}\hfill & \hfill \frac{1}{4}\hfill \\ \hfill \hfill \\ \hfill 1\hfill & \hfill \frac{1}{4}\hfill & \hfill \frac{1}{5}\hfill \end{array}\right]$ d) $D=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -6\hfill & \hfill 17\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 11\hfill & \hfill -8\hfill & \hfill 0\hfill \end{array}\right]$ e) none of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
We could check by carrying out matrix multiplication that none of $AB$, $AC$, $AD$ is $I$, the 3 by 3 identity matrix, so (2), (3) and (4) are all incorrect. (In particular, note that ${A}^{-1}$ does not consist of the inverses of the entries of $A$!!) This leaves (1) and (5), and to decide which of them is true, we can augment $A$ by the 3 by 3 identity and reduce: $\begin{array}{llll}\hfill \left[A|I\right]& =\left[\begin{array}{cccccc}\hfill -3\hfill & \hfill -1\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill -1\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -5\hfill & \hfill -6\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 11\hfill & \hfill 13\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6∕5\hfill & \hfill 0\hfill & \hfill -2∕5\hfill & \hfill 2∕5\hfill \\ \hfill 0\hfill & \hfill 11\hfill & \hfill 13\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]\to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6∕5\hfill & \hfill 0\hfill & \hfill -2∕5\hfill & \hfill 2∕5\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1∕5\hfill & \hfill 1\hfill & \hfill 22∕5\hfill & \hfill -7∕5\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ at which point it is clear that there will be a leading 1 in row 3, column 3, so ${A}^{-1}$ exists. Hence (5) is correct. We do not need to find ${A}^{-1}$, but continuing the row reduction yields $\left[I|{A}^{-1}\right]$, and in fact shows that ${A}^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 13\hfill & \hfill -8\hfill \\ \hfill -5\hfill & \hfill -11\hfill & \hfill 7\hfill \end{array}\right].$
Let $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 5\hfill & \hfill -7\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 12\hfill & \hfill -27\hfill \end{array}\right]$. The inverse of $A$ (i.e., ${A}^{-1}$) is: Exactly one option must be correct)
 a) undefined b) $B=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 5\hfill & \hfill -7\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 12\hfill & \hfill -27\hfill \end{array}\right]$ c) $C=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill \frac{1}{5}\hfill & \hfill -\frac{1}{7}\hfill \\ \hfill \hfill \\ \hfill \frac{1}{2}\hfill & \hfill \frac{1}{3}\hfill & \hfill \frac{1}{6}\hfill \\ \hfill \hfill \\ \hfill 1\hfill & \hfill \frac{1}{12}\hfill & \hfill -\frac{1}{27}\hfill \end{array}\right]$ d) $D=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -6\hfill & \hfill 17\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 11\hfill & \hfill -8\hfill & \hfill 0\hfill \end{array}\right]$ e) none of the above

Choice (a) is correct!
We could check by carrying out matrix multiplication that none of $AB$, $AC$, $AD$ is $I$, the 3 by 3 identity matrix, so (2), (3) and (4) are all incorrect. (In particular, note that ${A}^{-1}$ does not consist of the inverses of the entries of $A$!!) This leaves (1) and (5), and to decide which of them is true, we can augment $A$ by the 3 by 3 identity and reduce: $\begin{array}{llll}\hfill \left[A|I\right]& =\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 5\hfill & \hfill -7\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 12\hfill & \hfill -27\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 5\hfill & \hfill -7\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -7\hfill & \hfill 20\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 7\hfill & \hfill -20\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \to \left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 5\hfill & \hfill -7\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -7\hfill & \hfill 20\hfill & \hfill -2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ at which point it is clear that there can be no leading 1 in row 3, column 3, so (1) is correct.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Which of the following matrices are elementary matrices? (Zero or more options can be correct)
 a) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ b) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ c) $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ d) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

There is at least one mistake.
For example, choice (a) should be True.
This matrix corresponds to the elementary row operation ${R}_{2}↔{R}_{3}$.
There is at least one mistake.
For example, choice (b) should be False.
This matrix is the product of two elementary matrices corresponding to ${R}_{2}:=2{R}_{2}$ then ${R}_{2}:={R}_{2}+{R}_{3}$
There is at least one mistake.
For example, choice (c) should be False.
This matrix is product of two elementary matrices corresponding to ${R}_{1}↔{R}_{3}$ then ${R}_{2}↔{R}_{1}$
There is at least one mistake.
For example, choice (d) should be True.
This matrix to the elementary row operation ${R}_{1}:={R}_{1}+2{R}_{2}$.
Correct!
1. True This matrix corresponds to the elementary row operation ${R}_{2}↔{R}_{3}$.
2. False This matrix is the product of two elementary matrices corresponding to ${R}_{2}:=2{R}_{2}$ then ${R}_{2}:={R}_{2}+{R}_{3}$
3. False This matrix is product of two elementary matrices corresponding to ${R}_{1}↔{R}_{3}$ then ${R}_{2}↔{R}_{1}$
4. True This matrix to the elementary row operation ${R}_{1}:={R}_{1}+2{R}_{2}$.
The elementary matrix corresponding to the Elementary Row Operation ${R}_{3}={R}_{3}-5{R}_{2}$ on a matrix with 3 rows is: Exactly one option must be correct)
 a) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ b) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ c) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -5\hfill & \hfill 1\hfill \end{array}\right]$ d) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 1\hfill \end{array}\right]$ e) none of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The correct answer is (3), since that is the result of applying the give elementary row operation to the 3 by 3 identity matrix.
Note that $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -5\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e-5c\hfill & \hfill f-5d\hfill & \hfill \cdots \hfill \end{array}\right]$ is the matrix we would get by applying the given elementary row operation to $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \end{array}\right]$.
Choice (d) is incorrect
Choice (e) is incorrect
The elementary matrix corresponding to the Elementary Row Operation ${R}_{3}=\frac{1}{9}{R}_{3}$ on a matrix with 4 rows is: Exactly one option must be correct)
 a) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 9\hfill \end{array}\right]$ b) $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{9}\hfill \end{array}\right]$ c) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 9\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ d) $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{9}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ e) none of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The correct answer is (4), since that is the result of applying the give elementary row operation to the 4 by 4 identity matrix.
Note that $\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{9}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill \cdots \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill \frac{1}{9}e\hfill & \hfill \frac{1}{9}f\hfill & \hfill \cdots \hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill \cdots \hfill \end{array}\right]$ is the matrix we would get by applying the given elementary row operation to $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill \cdots \hfill \end{array}\right]$.
Choice (e) is incorrect
Consider the six elementary matrices below:
$\begin{array}{ccc}\hfill {E}_{1}=\left[\begin{array}{cccc}\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 & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right],\hfill & \hfill {E}_{2}=\left[\begin{array}{cccc}\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 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],\hfill & \hfill {E}_{3}=\left[\begin{array}{cccc}\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 & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right],\hfill \\ & & \\ \hfill {E}_{4}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\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 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right],\hfill & \hfill {E}_{5}=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -2\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 & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right],\hfill & \hfill {E}_{6}=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 2\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 & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right].\hfill \end{array}$
Which of the following statements are correct? (Zero or more options can be correct)
 a) The inverse of ${E}_{1}$ is ${E}_{2}$. b) The inverse of ${E}_{1}$ is ${E}_{3}$. c) The inverse of ${E}_{2}$ is ${E}_{2}$. d) The inverse of ${E}_{5}$ is ${E}_{6}$. e) The inverse of ${E}_{4}$ is ${E}_{4}$. f) The inverse of ${E}_{3}$ is ${E}_{6}$.

There is at least one mistake.
For example, choice (a) should be False.
${E}_{1}$ corresponds to the row operation ${R}_{4}={R}_{4}+2{R}_{2}$, whereas ${E}_{2}$ corresponds to ${R}_{1}↔{R}_{2}$.
There is at least one mistake.
For example, choice (b) should be True.
${E}_{1}$ corresponds to the row operation ${R}_{4}={R}_{4}+2{R}_{2}$ and ${E}_{3}$ corresponds to the row operation ${R}_{4}={R}_{4}-2{R}_{2}$.
There is at least one mistake.
For example, choice (c) should be True.
${E}_{2}$ corresponds to ${R}_{1}↔{R}_{2}$.
There is at least one mistake.
For example, choice (d) should be True.
${E}_{5}$ corresponds to the row operation ${R}_{1}={R}_{1}-2{R}_{2}$, whereas ${E}_{6}$ corresponds to ${R}_{1}={R}_{1}+2{R}_{2}$.
There is at least one mistake.
For example, choice (e) should be True.
${E}_{4}$ corresponds to ${R}_{1}↔{R}_{3}$.
There is at least one mistake.
For example, choice (f) should be False.
${E}_{3}$ corresponds to the row operation ${R}_{4}={R}_{4}-2{R}_{2}$, whereas ${E}_{6}$ corresponds to ${R}_{1}={R}_{1}+2{R}_{2}$.
Correct!
1. False ${E}_{1}$ corresponds to the row operation ${R}_{4}={R}_{4}+2{R}_{2}$, whereas ${E}_{2}$ corresponds to ${R}_{1}↔{R}_{2}$.
2. True ${E}_{1}$ corresponds to the row operation ${R}_{4}={R}_{4}+2{R}_{2}$ and ${E}_{3}$ corresponds to the row operation ${R}_{4}={R}_{4}-2{R}_{2}$.
3. True ${E}_{2}$ corresponds to ${R}_{1}↔{R}_{2}$.
4. True ${E}_{5}$ corresponds to the row operation ${R}_{1}={R}_{1}-2{R}_{2}$, whereas ${E}_{6}$ corresponds to ${R}_{1}={R}_{1}+2{R}_{2}$.
5. True ${E}_{4}$ corresponds to ${R}_{1}↔{R}_{3}$.
6. False ${E}_{3}$ corresponds to the row operation ${R}_{4}={R}_{4}-2{R}_{2}$, whereas ${E}_{6}$ corresponds to ${R}_{1}={R}_{1}+2{R}_{2}$.
Let $A=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$, then ${A}^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 8\hfill \\ \hfill -1\hfill & \hfill 3\hfill & \hfill -7\hfill \end{array}\right]$.
What sequence of elementary row operations is needed to transform $\left[A|I\right]$ to $\left[I|{A}^{-1}\right]$ ? Exactly one option must be correct)
 a) ${R}_{2}:={R}_{2}-\frac{1}{3}{R}_{1}$, ${R}_{2}:=3{R}_{2}$, ${R}_{2}↔{R}_{3}$, ${R}_{2}:={R}_{2}-{R}_{3},$${R}_{3}:={R}_{3}-7{R}_{2}$, ${R}_{1}:={R}_{1}+{R}_{2}$, ${R}_{1}:={R}_{1}-{R}_{3}$, ${R}_{1}:=\frac{1}{3}{R}_{1}$ b) ${R}_{1}:=\frac{1}{3}{R}_{1}$, ${R}_{2}:={R}_{2}-{R}_{1}$, ${R}_{2}:=\frac{3}{7}{R}_{2}$, ${R}_{3}:={R}_{3}-{R}_{2}$${R}_{3}:=-7{R}_{3}$, ${R}_{2}:={R}_{2}-\frac{8}{7}{R}_{3}$, ${R}_{1}:={R}_{1}+\frac{1}{3}{R}_{2}$, ${R}_{1}:={R}_{1}-\frac{1}{3}{R}_{3}$ c) ${R}_{2}:={R}_{2}-\frac{1}{3}{R}_{1}$, ${R}_{3}:={R}_{3}-{R}_{2}$, ${R}_{2}:=\frac{3}{7}{R}_{2}$, ${R}_{3}:=-7{R}_{3},$${R}_{2}:={R}_{1}-{R}_{3}$, ${R}_{1}:={R}_{1}+{R}_{2}$, ${R}_{1}:={R}_{1}-{R}_{3}$ d) All of the above sequences

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
The system of equations $\begin{array}{rcll}3x-y+z& =& 1,& \text{}\\ x+2y+3z& =& 1,& \text{}\\ y+z& =& -1,& \text{}\end{array}$ can be written in the form $A\mathbf{x}=\mathbf{b}$ where
$A=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right],\mathbf{x}=\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\mathbf{b}=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right].$
Using ${A}^{-1}$ (given in question 8) to solve the system of equations, which of the following statements is correct ? Exactly one option must be correct)
 a) $y=-10$ b) $z=-5$ c) $x=6$ d) None of the above

Choice (a) is correct!
${A}^{-1}\mathbf{b}=\left[\begin{array}{c}\hfill -6\hfill \\ \hfill -10\hfill \\ \hfill 9\hfill \end{array}\right]⇒x=-6$, $y=-10$, $z=9$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Let $A=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$, and suppose ${A}^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 6\hfill & \hfill -8\hfill \end{array}\right]$. Consider the system $\begin{array}{llll}\hfill ax+by+cz& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill dx+ey+fz& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill gx+hy+iz& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Which of the following is true ? Exactly one option must be correct)
 a) the system has no solutions; b) the system has many solutions; c) the system has the unique solution $x=16,\phantom{\rule{2.77695pt}{0ex}}y=23,\phantom{\rule{2.77695pt}{0ex}}z=-21$; d) the system has the unique solution $x=14,\phantom{\rule{2.77695pt}{0ex}}y=-22,\phantom{\rule{2.77695pt}{0ex}}z=13$; e) none of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Since $A$ is invertible, there is the unique solution $\begin{array}{llll}\hfill \left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right]& ={A}^{-1}\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 6\hfill & \hfill -8\hfill \end{array}\right]\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 1\hfill \\ \hfill 4\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left[\begin{array}{c}\hfill 16\hfill \\ \hfill 23\hfill \\ \hfill -21\hfill \end{array}\right].\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ Hence the correct answer is (3).
Choice (d) is incorrect
Choice (e) is incorrect