# Quiz 8: Inverses and elementary matrices

Question

## Question 1

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:
 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 invertable by aplying 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 $A\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]={I}_{2}=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]A$.
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 invertable by aplying 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 $A\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]={I}_{2}=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right]A$.

## Question 2

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:
 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

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
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 ${A}^{-1}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 13\hfill & \hfill -8\hfill \\ \hfill -5\hfill & \hfill -11\hfill & \hfill 7\hfill \end{array}\right].$

## Question 3

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:
 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

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.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 4

Which of the following matrices are elementary matrices?
 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 ${R}_{1}:={R}_{1}+2{R}_{2}$.
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 ${R}_{1}:={R}_{1}+2{R}_{2}$.

## Question 5

The elementary matrix corresponding to the Elementary Row Operation ${R}_{3}={R}_{3}-5{R}_{2}$ on a matrix with 3 rows is:
 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

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
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 $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \end{array}\right]$.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.

## Question 6

The elementary matrix corresponding to the Elementary Row Operation ${R}_{3}=\frac{1}{9}{R}_{3}$ on a matrix with 4 rows is:
 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

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
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 $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill \cdots \hfill \\ \hfill c\hfill & \hfill d\hfill & \hfill \cdots \hfill \\ \hfill e\hfill & \hfill f\hfill & \hfill \cdots \hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill \cdots \hfill \end{array}\right]$.
Not correct. Choice (e) is false.

## Question 7

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?
 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 ${R}_{1}={R}_{1}+2{R}_{2}$.
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 ${R}_{1}={R}_{1}+2{R}_{2}$.

## Question 8

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 $\left[I|{A}^{-1}\right]$ ?
 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

Not correct. Choice (a) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 9

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 $Ax=b$ where
$A=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right],x=\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \\ \hfill z\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}b=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right].$
Using ${A}^{-1}$ (given in question 8) to solve the system of equations, which of the following statements is correct ?
 a) $y=-10$ b) $z=-5$ c) $x=6$ d) None of the above

${A}^{-1}b=\left[\begin{array}{c}\hfill -6\hfill \\ \hfill -10\hfill \\ \hfill 9\hfill \end{array}\right]⇒x=-6$, $y=-10$, $z=9$.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 10

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 ?
 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.

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
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).