Question

1-7 lesson quizsolving linear systems using matrices1. which of the following operations would not create an equivalent matrix?(a) switching two rows(b) adding two rows together and replacing one of those rows with the result(c) adding a constant to every element of a row(d) multiplying every element of a row by a constant2. what is the solution to the linear system represented by the matrix?$\begin{bmatrix} 1 & 0 & mid & -4\\ 0 & 1 & mid & 13end{bmatrix}$(a) (1, 0, -4)(b) (0, 1, 13)(c) (1, 1)(d) (-4, 13)3. use matrices to solve the system of equations.$\begin{cases}x - 3y + 3z = -8 \\7x + y + 2z = 6 \\3x + 2y - z = 4end{cases}$4. philip made a total of 9 bracelets and necklaces from 120 inches of cord. he used 8 inches of cord for each bracelet and 20 inches of cord for each necklace. use matrices to find the number of bracelets and the number of necklaces that philip made.5. at the movies, devin bought three tickets, one drink, and one bag of popcorn for $43.75. neil bought one ticket, three drinks, and two bags of popcorn for $27.75. jung bought two tickets, two drinks, and one bag of popcorn for $34. make an augmented matrix to represent the situation and use it to find out the prices for one ticket, one drink, and one bag of popcorn.

Explanation:

Question 1

Step1: Recall row equivalence rules

Elementary row operations that produce equivalent matrices are: swapping rows, adding a multiple of one row to another, multiplying a row by a non-zero constant. Adding a constant to every element of a row is not an elementary row operation and does not preserve equivalence.

Step2: Identify incorrect operation

Option C does not follow valid row equivalence rules.

Question 2

Step1: Interpret augmented matrix

The matrix $\begin{bmatrix}1 & 0 & \mid & -4\\0 & 1 & \mid & 13\end{bmatrix}$ corresponds to the system:
$1x + 0y = -4$
$0x + 1y = 13$

Step2: Extract solution

This gives $x=-4$, $y=13$, so the solution is $(-4,13)$.

Question 3

Step1: Write augmented matrix

The system $\begin{cases}x - 3y + 3z = -8\\7x + y + 2z = 6\\3x + 2y - z = 4\end{cases}$ becomes:
$$\begin{bmatrix}1 & -3 & 3 & \mid & -8\\7 & 1 & 2 & \mid & 6\\3 & 2 & -1 & \mid & 4\end{bmatrix}$$

Step2: Eliminate $x$ from row2/3

Row2 = Row2 - 7*Row1: $7-7(1)=0$, $1-7(-3)=22$, $2-7(3)=-19$, $6-7(-8)=62$
Row3 = Row3 - 3*Row1: $3-3(1)=0$, $2-3(-3)=11$, $-1-3(3)=-10$, $4-3(-8)=28$
Matrix:
$$\begin{bmatrix}1 & -3 & 3 & \mid & -8\\0 & 22 & -19 & \mid & 62\\0 & 11 & -10 & \mid & 28\end{bmatrix}$$

Step3: Simplify row2/3

Row2 = $\frac{1}{22}$Row2:
$$\begin{bmatrix}1 & -3 & 3 & \mid & -8\\0 & 1 & -\frac{19}{22} & \mid & \frac{31}{11}\\0 & 11 & -10 & \mid & 28\end{bmatrix}$$
Row3 = Row3 - 11*Row2: $11-11(1)=0$, $-10-11(-\frac{19}{22})=-\frac{1}{2}$, $28-11(\frac{31}{11})=-3$
Matrix:
$$\begin{bmatrix}1 & -3 & 3 & \mid & -8\\0 & 1 & -\frac{19}{22} & \mid & \frac{31}{11}\\0 & 0 & -\frac{1}{2} & \mid & -3\end{bmatrix}$$

Step4: Solve for $z$

Row3: $-\frac{1}{2}z=-3 \implies z=6$

Step5: Solve for $y$

Substitute $z=6$ into Row2: $y - \frac{19}{22}(6)=\frac{31}{11}$
$y = \frac{31}{11} + \frac{114}{22} = \frac{62+114}{22} = \frac{176}{22}=8$

Step6: Solve for $x$

Substitute $y=8,z=6$ into Row1: $x -3(8)+3(6)=-8$
$x -24+18=-8 \implies x=-2$

Question 4

Step1: Define variables & set up system

Let $b$ = number of bracelets, $n$ = number of necklaces.
$\begin{cases}b + n = 9\\8b + 20n = 120\end{cases}$

Step2: Write augmented matrix

$$\begin{bmatrix}1 & 1 & \mid & 9\\8 & 20 & \mid & 120\end{bmatrix}$$

Step3: Eliminate $b$ from row2

Row2 = Row2 -8*Row1: $8-8(1)=0$, $20-8(1)=12$, $120-8(9)=48$
Matrix:
$$\begin{bmatrix}1 & 1 & \mid & 9\\0 & 12 & \mid & 48\end{bmatrix}$$

Step4: Solve for $n$

Row2: $12n=48 \implies n=4$

Step5: Solve for $b$

Substitute $n=4$ into Row1: $b +4=9 \implies b=5$

Question 5

Step1: Define variables & set up system

Let $t$ = ticket price, $d$ = drink price, $p$ = popcorn price.
$\begin{cases}3t + d + p = 43.75\\t + 3d + 2p = 27.75\\2t + 2d + p = 34\end{cases}$

Step2: Write augmented matrix

$$\begin{bmatrix}3 & 1 & 1 & \mid & 43.75\\1 & 3 & 2 & \mid & 27.75\\2 & 2 & 1 & \mid & 34\end{bmatrix}$$

Step3: Swap row1 and row2

$$\begin{bmatrix}1 & 3 & 2 & \mid & 27.75\\3 & 1 & 1 & \mid & 43.75\\2 & 2 & 1 & \mid & 34\end{bmatrix}$$

Step4: Eliminate $t$ from row2/3

Row2 = Row2 -3*Row1: $3-3(1)=0$, $1-3(3)=-8$, $1-3(2)=-5$, $43.75-3(27.75)=-39.5$
Row3 = Row3 -2*Row1: $2-2(1)=0$, $2-2(3)=-4$, $1-2(2)=-3$, $34-2(27.75)=-21.5$
Matrix:
$$\begin{bmatrix}1 & 3 & 2 & \mid & 27.75\\0 & -8 & -5 & \mid & -39.5\\0 & -4 & -3 & \mid & -21.5\end{bmatrix}$$

Step5: Simplify row2/3

Row2 = $\frac{1}{-8}$Row2:
$$\begin{bmatrix}1 & 3 & 2 & \mid & 27.75\\0 & 1 & \frac{5}{8} & \mid & 4.9375\\0 & -4 & -3 & \mid & -21.5\end{bmatrix}$$
Row3 = Row3 +4*Row2: $-4+4(1)=0$, $-3+4(\frac{5}{8})=-\frac{1}{2}$, $-21.5+4(4.9375)=-1.75$
Matrix:
$$\begin{bmatrix}1 & 3 & 2 & \mid & 27.75\\0 & 1 & \frac{5}{8} & \mid & 4.9375\\0 & 0 & -\frac{1}{2} & \mid & -1.75\end{bmatrix}$$

Step6: Solve for $p$

Row3: $-\frac{1}{2}p=-1.75 \implies p=3.50$

Step7: Solve for $d$

Substitute $p=3.50$ into Row2: $d + \frac{5}{8}(3.50)=4.9375$
$d = 4.9375 - 2.1875 = 2.75$

Step8: Solve for $t$

Substitute $d=2.75,p=3.50$ into Row1: $t +3(2.75)+2(3.50)=27.75$
$t +8.25+7=27.75 \implies t=12.50$

Answer:

1. C. Adding a constant to every element of a row
2. D. (-4, 13)
3. $x=-2$, $y=8$, $z=6$
4. 5 bracelets, 4 necklaces
5. Ticket: $\$12.50$, Drink: $\$2.75$, Popcorn: $\$3.50$