Question 1

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{array} { l } 
2 ( x - 2 z ) = 3 y + x + 30 \
x = 2 y - 2 z - 3 \
- 5 x + 2 y + 6 z = - 41
\end{array}$$
A) $\{ ( 3 , - 1,5 ) \}$
B) $\{ ( 1 , - 3 , - 5 ) \}$
C) $\{ ( - 3,1 , - 5 ) \}$
D) $\{ ( - 1,3,5 ) \}$

Accepted Answer

To solve the system using Gaussian elimination or Gauss-Jordan elimination, we first write the equations in standard form:1. $2x - 3y - 4z = 30$2. $x - 2y + 2z = -3$3. $-5x + 2y + 6z = -41$Next, we form the augmented matrix and apply row operations to reach row-echelon form or reduced row-echelon form.The augmented matrix is:$$\begin{bmatrix}2 & -3 & -4 & | & 30 \1 & -2 & 2 & | & -3 \-5 & 2 & 6 & | & -41 \\end{bmatrix}$$Applying row operations to simplify:1. $R2 \rightarrow R2 - \frac{1}{2}R1$2. $R3 \rightarrow R3 + \frac{5}{2}R1$3. And so on, to isolate variables.After simplification, we find the solution:$x = 1$, $y = -3$, $z = -5$.Thus, the correct answer is $\{ ( 1 , - 3 , - 5 ) \}$.

Question 2

Perform the elementary row operation on the given matrix.
-$\begin{array} { l } 
R _ { 2 } \Leftrightarrow R _ { 3 } \
{ \left[ \begin{array} { r r r | r } 
- 8 & 2 & - 9 & - 4 \
- 3 & 7 & - 6 & 6 \
5 & 0 & 3 & 6
\end{array} \right] }
\end{array}$

A) $ \left[\begin{array}{rrr|r}-8 & 2 & -9 & -4 \ 2 & 7 & -3 & 9 \ 5 & 0 & 3 & 6\end{array}\right] $

B) $ \left[\begin{array}{rrr|r}-8 & 2 & -9 & -4 \ 5 & 0 & 3 & 6 \ -3 & 7 & -6 & 6\end{array}\right] $

C) $ \left[\begin{array}{rrr|r}-3 & 7 & -6 & 6 \ -8 & 2 & -9 & -4 \ 5 & 0 & 3 & 6\end{array}\right] $

D) $ \left[\begin{array}{rrr|r}-8 & 2 & -9 & -4 \ -3 & 7 & -6 & 6 \ 2 & 7 & -3 & 9\end{array}\right] $

Accepted Answer

The operation $R_2 \Leftrightarrow R_3$ means to swap row 2 with row 3. Choice B correctly shows the result of this operation, with the original row 2 (-3, 7, -6, 6) and row 3 (5, 0, 3, 6) swapped.

Question 3

Perform the indicated row operations, then write the new matrix.
-$$\left[ \begin{array} { r r r | r } 
1 & 1 & 1 & - 1 \
- 2 & 3 & 5 & 3 \
3 & 2 & 4 & 1
\end{array} \right] \begin{array} { c } 
2 R 1 + R 2 \rightarrow R 2 , \
- 3 R 1 + R 3 \rightarrow R 3
\end{array}$$
A)
$$\left[ \begin{array} { r r r | r } 
1 & 1 & 1 & - 1 \
0 & 5 & 7 & 1 \
0 & - 1 & 1 & 4
\end{array} \right]$$
B)
$$\left[ \begin{array} { r r r | r } 
1 & 1 & 1 & - 1 \
0 & 5 & 7 & 3 \
0 & - 1 & 1 & 1
\end{array} \right]$$
C)$$\left[ \begin{array} { r r r | r } 
1 & 1 & 1 & - 1 \
- 2 & 3 & 5 & 3 \
0 & - 1 & 1 & 1
\end{array} \right]$$
D)
$\left[ \begin{array} { l l l | r } 1 & 1 & 1 & - 1 \ 0 & 5 & 7 & 1 \ 3 & 2 & 4 & 1 \end{array} \right]$

Accepted Answer

For $2R_1 + R_2$, we calculate $2(1, 1, 1, -1) + (-2, 3, 5, 3) = (0, 5, 7, 1)$ for the second row. For $-3R_1 + R_3$, we calculate $-3(1, 1, 1, -1) + (3, 2, 4, 1) = (0, -1, 1, 4)$ for the third row.

Question 4

Determine if the matrix is in row-echelon form.
-$$\left[ \begin{array} { r r r | r } 
1 & - 6 & - 8 & - 7 \
0 & 1 & 0 & 8 \
0 & 0 & 2 & 3
\end{array} \right]$$
A) No
B) Yes

Accepted Answer

The answer of Determine if the matrix is in row-echelon...

Question 5

Write the word or phrase that best completes each statement or answers the question.
Provide the missing information.
-True or false? A system of linear equations in three variables may have exactly two solutions.

Accepted Answer

A system of linear equations in three variables can have either one solution, no solution, or infinitely many solutions, but it cannot have exactly two solutions.

Question 6

Find the partial fraction decomposition for the given rational expression. Use the technique of Gaussian
elimination to find A, B, and C.
-$$\frac { - 3 x ^ { 2 } + 7 x - 10 } { ( x + 2 ) ( x - 1 ) ^ { 2 } } = \frac { A } { x + 2 } + \frac { B } { x - 1 } - \frac { C } { ( x - 1 ) ^ { 2 } }$$
A) $\frac { - 4 } { x + 2 } + \frac { 1 } { x - 1 } - \frac { 2 } { ( x - 1 ) ^ { 2 } }$
B) $\frac { - 4 } { x + 2 } + \frac { 2 } { x - 1 } - \frac { 2 } { ( x - 1 ) ^ { 2 } }$
C) $\frac { 4 } { x + 2 } + \frac { 1 } { x - 1 } - \frac { 2 } { ( x - 1 ) ^ { 2 } }$
D) $\frac { - 4 } { x + 2 } + \frac { 1 } { x - 1 } + \frac { 2 } { ( x - 1 ) ^ { 2 } }$

Accepted Answer

The correct partial fraction decomposition is $\frac { - 4 } { x + 2 } + \frac { 1 } { x - 1 } - \frac { 2 } { ( x - 1 ) ^ { 2 } }$ because when you combine these fractions over a common denominator and simplify, you get the original expression.

Question 7

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{array} { l } 
- 5 y = 24 - 3 x \
2 ( x - 4 y ) = 49 - y
\end{array}$$
A) $\{ ( 7,9 ) \}$
B) $\{ ( - 7 , - 9 ) \}$
C) $\{ ( 9,7 ) \}$
D) $\{ ( - 9 , - 7 ) \}$

Accepted Answer

First, simplify and rearrange the equations into standard form (Ax + By = C). The first equation is already given as $-3x - 5y = 24$. The second equation simplifies to $2x - 8y = 49 - y$, which rearranges to $2x - 7y = 49$. Then, using either Gaussian elimination or Gauss-Jordan elimination, you can solve the system to find $x = -7$ and $y = -9$, which corresponds to choice B.

Question 8

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{array} { l } 
5 x + 9 z = - 16 + 6 y \
8 x = 7 y + 7 z - 141 \
8 z = - 2 y - 2 x + 42
\end{array}$$
A) $\{ ( 5 , - 8,6 ) \}$
B) $\{ ( 5,8 , - 6 ) \}$
C) $\{ ( 8 , - 5,6 ) \}$
D) $\{ ( - 8,5,6 ) \}$

Accepted Answer

To solve the system using Gaussian elimination or Gauss-Jordan elimination, we first write the system in matrix form as follows:$$\begin{align*}5x + 0y + 9z &= -16 \8x - 7y - 7z &= -141 \-2x - 2y + 8z &= 42\end{align*}$$Next, we perform row operations to get the matrix into reduced row echelon form (RREF). The specific steps can vary, but the goal is to get leading 1s down the diagonal and zeros everywhere else in the matrix, if possible.After performing the necessary row operations (which are not detailed here due to the nature of the question), we solve for $x$, $y$, and $z$.The correct solution is $(x, y, z) = (-8, 5, 6)$, which matches choice D. This indicates that after performing Gaussian elimination or Gauss-Jordan elimination on the system of equations, the solution set is $\{(-8, 5, 6)\}$.

Question 9

Use a calculator to approximate the reduced row-echelon form of the augmented matrix representing the
given system. Give the solution set where x, y, and z are rounded to 2 decimal places.
-$\begin{aligned}
0.52 x - 3.79 y - 4.67 z & = 9.15 \
0.03 x + 0.06 y + 0.13 z & = 0.53 \
0.974 x + 0.813 y + 0.419 z & = 0.189
\end{aligned}$
A) $\{ ( 6.18 , - 11.2,7.82 ) \}$
B) $\{ ( - 15.61 , - 1.39 , - 2.84 ) \}$
C) $\{ ( 7.91 , - 1.03,3.18 ) \}$
D) $\{ ( - 4.49 , - 6.63,2.49 ) \}$

Accepted Answer

The reduced row-echelon form of the augmented matrix is:$$\begin{bmatrix} 1 & 0 & 0 & 6.18 \ 0 & 1 & 0 & -11.2 \ 0 & 0 & 1 & 7.82 \end{bmatrix}$$So the solution set is:$$\{(6.18,-11.2,7.82)\}$$

Question 10

Perform the elementary row operation on the given matrix.
-$- 5 R _ { 1 } + R _ { 3 } \rightarrow R _ { 3 }$
$\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \ 7 & 18 & 5 & 8 \ 5 & 12 & 9 & 13 \end{array} \right]$
A) $\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \ 2 & - 62 & - 20 & - 7 \ 5 & 12 & 9 & 13 \end{array} \right]$ 
B) $\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \ 7 & 18 & 5 & 8 \ 0 & - 68 & - 16 & - 2 \end{array} \right]$ 
C) $\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \ 7 & 18 & 5 & 8 \ 0 & - 68 & - 16 & 16 \end{array} \right]$
D) $\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \ 7 & 18 & 5 & 8 \ 6 & 28 & 14 & 16 \end{array} \right]$

Accepted Answer

To perform the operation $-5R_1 + R_3 \rightarrow R_3$, we multiply each element of row 1 by -5 and add the corresponding element from row 3, then replace row 3 with the result. For each column: - Column 1: $(-5 \times 1) + 5 = 0$- Column 2: $(-5 \times 16) + 12 = -80 + 12 = -68$- Column 3: $(-5 \times 5) + 9 = -25 + 9 = -16$- Column 4 (right side of the augmented matrix): $(-5 \times 3) + 13 = -15 + 13 = -2$This matches option B.

Question 11

Determine if the matrix is in row-echelon form.
-$$\left[ \begin{array} { r r r r | r } 
1 & 0 & 0 & 0 & 2 \
0 & 1 & 0 & 0 & 4 \
0 & 0 & - 1 & 0 & 9 \
0 & 0 & 0 & 1 & - 6
\end{array} \right]$$
A) No
B) Yes

Accepted Answer

The answer of Determine if the matrix is in row-echelon...

Question 12

Solve the problem.
-Danielle stayed in three different cities (Washington, D.C., Atlanta, Georgia, and Dallas, Texas) for a total of 22 nights. She spent twice as many nights in Dallas as she did in Washington. The total
Cost for 22 nights (excluding tax) was $3,100. Determine the number of nights that she spent in each city. $$\begin{array} { l | c }  { \text { City } } & \text { Cost per Night } \
\hline \text { Washington } & \$ 100 \
\hline \text { Atlanta } & \$ 175 \
\hline \text { Dallas } & \$ 150
\end{array}$$
A) 5 nights in Washington, 7 nights in Atlanta, and 10 nights in Dallas
B) 1 night in Washington, 19 nights in Atlanta, and 2 nights in Dallas
C) 4 nights in Washington, 10 nights in Atlanta, and 8 nights in Dallas
D) 6 nights in Washington, 4 nights in Atlanta, and 12 nights in Dallas

Accepted Answer

The answer of Solve the problem.
-Danielle stayed in three different...

Question 13

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{aligned}
5 x - 7 y & = - 3 \
- 6 x + 5 y & = 7
\end{aligned}$$
A) $\{ ( - 2 , - 1 ) \}$
B) $\{ ( - 1 , - 2 ) \}$
C) $\{ ( 2,1 ) \}$
D) $\{ ( - 1,2 ) \}$

Accepted Answer

To solve the system using Gaussian elimination, we first write the system in matrix form:$$\begin{align*}\begin{bmatrix}5 & -7 \-6 & 5 \\end{bmatrix}\begin{bmatrix}x \y \\end{bmatrix}=\begin{bmatrix}-3 \7 \\end{bmatrix}\end{align*}$$Step 1: Make the leading coefficient of the first row 1 (if it's not already). In this case, it's not necessary since we're focusing on elimination.Step 2: Eliminate the x-term from the second equation. We can multiply the first row by 6 and the second row by 5 and then add them together to eliminate x:$$\begin{align*}6(5x - 7y) &= 6(-3) \5(-6x + 5y) &= 5(7) \\end{align*}$$This gives us:$$\begin{align*}30x - 42y &= -18 \-30x + 25y &= 35 \\end{align*}$$Adding these equations:$$-17y = 17$$Solving for y:$$y = -1$$Step 3: Substitute y back into one of the original equations to solve for x. Using the first equation:$$5x - 7(-1) = -3$$$$5x + 7 = -3$$$$5x = -10$$$$x = -2$$Therefore, the solution is $\{ (-2, -1) \}$.

Question 14

Write the word or phrase that best completes each statement or answers the question.
Provide the missing information.
-True or false? A system of linear equations in three variables may have exactly one solution.

Accepted Answer

A system of linear equations in three variables can have exactly one solution if the equations represent three planes that intersect at a single point in three-dimensional space.

Question 15

Write the word or phrase that best completes each statement or answers the question.
Provide the missing information.
-True or false? A system of linear equations in three variables may have no solution.

Accepted Answer

A system of linear equations in three variables can be inconsistent, meaning the equations represent planes that do not all intersect at a single point, which results in no solution.

Question 16

Solve the problem.
-Andre borrowed $40,000 to buy a truck for his business. He borrowed from his parents who charge him 3% simple interest. He borrowed from a credit union that charges 4% simple interest, and he
Borrowed from a bank that charges 6% simple interest. He borrowed four times as much from his
Parents as from the bank, and the amount of interest he paid at the end of 1 yr was $1,560. How
Much did he borrow from each source?
A) He borrowed $4,000 from his parents, $35,000 from the credit union, and $1,000 from the bank.
B) He borrowed 1,500 from his parents, 32,500 from the credit union, and 6,000 from the bank.
C) He borrowed 6,000 from his parents, 32,500 from the credit union, and 1,500 from the bank.
D) He borrowed $8,000 from his parents, $30,000 from the credit union, and $2,000 from the bank.

Accepted Answer

The total amount borrowed is $40,000. Let x be the amount borrowed from the bank. Then, 4x is the amount borrowed from his parents, and the remaining $40,000 - 4x - x = $40,000 - 5x is borrowed from the credit union. The total interest paid is $1,560, which is the sum of the interests from all sources: 0.03(4x) + 0.04($40,000 - 5x) + 0.06x = $1,560. Solving this equation gives x = $2,000, so he borrowed $8,000 from his parents, $30,000 from the credit union, and $2,000 from the bank.

Question 17

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{aligned}
x + 4 y & = - 7 \
4 x + 3 y & = 11
\end{aligned}$$
A) $\{ ( 5 , - 3 ) \}$
B) $\left\{ \left( \frac { 17 } { 19 } , \frac { 65 } { 19 } \right) \right\}$
C) $\{ ( - 3,5 ) \}$
D) $\left\{ \left\{ - \frac { 65 } { 19 } , - \frac { 17 } { 19 } \right) \right\}$

Accepted Answer

The correct answer is A because the system has one solution, which is (5, -3).

Question 18

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$\begin{array} { l } 
x _ { 1 } - 5 x _ { 2 } + 2 x _ { 3 } = 6 \
5 x _ { 2 } - 2 x _ { 3 } = - 1 \
- 3 x _ { 2 } - 4 x _ { 3 } - 5 x _ { 4 } = - 30 \
5 x _ { 1 } + 4 x _ { 2 } + 4 x _ { 3 } + 5 x _ { 4 } = 56
\end{array}$

A) $\{ ( 5,1 , - 3 , - 3 ) \}$
B) $\{ ( - 5 , - 1 , - 3 , - 3 ) \}$
C) $\{ ( - 5 , - 1,3,3 ) \}$
D) $\{ ( 5,1,3,3 ) \}$

Accepted Answer

To solve the system using Gaussian elimination or Gauss-Jordan elimination, we start by writing the augmented matrix and then perform row operations to get it into reduced row echelon form (RREF). The given system is:1. $x_1 - 5x_2 + 2x_3 = 6$2. $5x_2 - 2x_3 = -1$3. $-3x_2 - 4x_3 - 5x_4 = -30$4. $5x_1 + 4x_2 + 4x_3 + 5x_4 = 56$The augmented matrix is:$$\left[\begin{array}{cccc|c}1 & -5 & 2 & 0 & 6 \0 & 5 & -2 & 0 & -1 \0 & -3 & -4 & -5 & -30 \5 & 4 & 4 & 5 & 56\end{array}\right]$$Through a series of row operations (which are not shown here for brevity), we aim to transform this matrix into RREF. The goal is to have leading 1s in each row, with zeros below and above each leading 1, and each leading 1 moves one column to the right as we move down the rows.After performing the necessary row operations, we can solve for $x_1$, $x_2$, $x_3$, and $x_4$. The solution that matches the operations correctly is $(5, 1, 3, 3)$, indicating that each variable corresponds to these values respectively. This process involves several steps, including scaling rows, adding and subtracting multiples of rows, and swapping rows to achieve the desired form. The exact operations depend on the specific strategy and steps taken, but the end result should yield the solution provided in option D.

Question 19

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{array} { l } 
w - 2 x + 3 y = 8 \
- 5 x - 5 y = - 25 \
- 2 x + 3 y - 4 z = 17 \
- 3 w - 5 x + 2 y - 3 z = - 4
\end{array}$$
A) $\{ ( - 3 , - 2,3 , - 3 ) \}$
B) $\{ ( - 3 , - 2 , - 3,3 ) \}$
C) $\{ ( 3,2,3 , - 3 ) \}$
D) $\{ ( 3,2 , - 3,3 ) \}$

Accepted Answer

The answer of Solve the system using Gaussian elimination or...

Question 20

Solve the system using Gaussian elimination or Gauss-Jordan elimination.
-$$\begin{array} { l } 
- 5 x + 9 y - 9 z = - 106 \
2 x - 2 y - 5 z = - 14 \
- 5 x - 8 y + 2 z = 11
\end{array}$$
A) $\{ ( - 3 , - 5 , - 6 ) \}$
B) $\{ ( - 3,5,6 ) \}$
C) $\{ ( - 5,3,6 ) \}$
D) $\{ ( 5 , - 3,6 ) \}$

Accepted Answer

The answer of Solve the system using Gaussian elimination or...

Quiz 6: Matrices and Determinants and Applications

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } 2 ( x - 2 z ) = 3 y + x + 30 \\x = 2 y - 2 z - 3 \\- 5 x + 2 y + 6 z = - 41\end{array}$

Perform the elementary row operation on the given matrix. - $\begin{array} { l } R _ { 2 } \Leftrightarrow R _ { 3 } \\{ \left[ \begin{array} { r r r | r } - 8 & 2 & - 9 & - 4 \\- 3 & 7 & - 6 & 6 \\5 & 0 & 3 & 6\end{array} \right] }\end{array}$

Perform the indicated row operations, then write the new matrix. - $\left[ \begin{array} { r r r | r } 1 & 1 & 1 & - 1 \\- 2 & 3 & 5 & 3 \\3 & 2 & 4 & 1\end{array} \right] \begin{array} { c } 2 R 1 + R 2 \rightarrow R 2 , \\- 3 R 1 + R 3 \rightarrow R 3\end{array}$

Determine if the matrix is in row-echelon form. - $\left[ \begin{array} { r r r | r } 1 & - 6 & - 8 & - 7 \\0 & 1 & 0 & 8 \\0 & 0 & 2 & 3\end{array} \right]$

Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -True or false? A system of linear equations in three variables may have exactly two solutions.

Find the partial fraction decomposition for the given rational expression. Use the technique of Gaussian elimination to find A, B, and C. - $\frac { - 3 x ^ { 2 } + 7 x - 10 } { ( x + 2 ) ( x - 1 ) ^ { 2 } } = \frac { A } { x + 2 } + \frac { B } { x - 1 } - \frac { C } { ( x - 1 ) ^ { 2 } }$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } - 5 y = 24 - 3 x \\2 ( x - 4 y ) = 49 - y\end{array}$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } 5 x + 9 z = - 16 + 6 y \\8 x = 7 y + 7 z - 141 \\8 z = - 2 y - 2 x + 42\end{array}$

Perform the elementary row operation on the given matrix. - $- 5 R _ { 1 } + R _ { 3 } \rightarrow R _ { 3 }$ $\left[ \begin{array} { r r r | r } 1 & 16 & 5 & 3 \\ 7 & 18 & 5 & 8 \\ 5 & 12 & 9 & 13 \end{array} \right]$

Determine if the matrix is in row-echelon form. - $\left[ \begin{array} { r r r r | r } 1 & 0 & 0 & 0 & 2 \\0 & 1 & 0 & 0 & 4 \\0 & 0 & - 1 & 0 & 9 \\0 & 0 & 0 & 1 & - 6\end{array} \right]$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{aligned}5 x - 7 y & = - 3 \\- 6 x + 5 y & = 7\end{aligned}$

Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -True or false? A system of linear equations in three variables may have exactly one solution.

Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -True or false? A system of linear equations in three variables may have no solution.

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{aligned}x + 4 y & = - 7 \\4 x + 3 y & = 11\end{aligned}$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } x _ { 1 } - 5 x _ { 2 } + 2 x _ { 3 } = 6 \\5 x _ { 2 } - 2 x _ { 3 } = - 1 \\- 3 x _ { 2 } - 4 x _ { 3 } - 5 x _ { 4 } = - 30 \\5 x _ { 1 } + 4 x _ { 2 } + 4 x _ { 3 } + 5 x _ { 4 } = 56\end{array}$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } w - 2 x + 3 y = 8 \\- 5 x - 5 y = - 25 \\- 2 x + 3 y - 4 z = 17 \\- 3 w - 5 x + 2 y - 3 z = - 4\end{array}$

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - $\begin{array} { l } - 5 x + 9 y - 9 z = - 106 \\2 x - 2 y - 5 z = - 14 \\- 5 x - 8 y + 2 z = 11\end{array}$

Quiz 6: Matrices and Determinants and Applications

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - 2(x−2z) =3y+x+30x=2y−2z−3−5x+2y+6z=−41\begin{array} { l } 2 ( x - 2 z ) = 3 y + x + 30 \\x = 2 y - 2 z - 3 \\- 5 x + 2 y + 6 z = - 41\end{array}2(x−2z) =3y+x+30x=2y−2z−3−5x+2y+6z=−41​

Determine if the matrix is in row-echelon form. - [1−6−8−701080023]\left[ \begin{array} { r r r | r } 1 & - 6 & - 8 & - 7 \\0 & 1 & 0 & 8 \\0 & 0 & 2 & 3\end{array} \right]​100​−610​−802​−783​​

Write the word or phrase that best completes each statement or answers the question. Provide the missing information. -True or false? A system of linear equations in three variables may have exactly two solutions.

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - −5y=24−3x2(x−4y) =49−y\begin{array} { l } - 5 y = 24 - 3 x \\2 ( x - 4 y ) = 49 - y\end{array}−5y=24−3x2(x−4y) =49−y​

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - 5x+9z=−16+6y8x=7y+7z−1418z=−2y−2x+42\begin{array} { l } 5 x + 9 z = - 16 + 6 y \\8 x = 7 y + 7 z - 141 \\8 z = - 2 y - 2 x + 42\end{array}5x+9z=−16+6y8x=7y+7z−1418z=−2y−2x+42​

Determine if the matrix is in row-echelon form. - [100020100400−1090001−6]\left[ \begin{array} { r r r r | r } 1 & 0 & 0 & 0 & 2 \\0 & 1 & 0 & 0 & 4 \\0 & 0 & - 1 & 0 & 9 \\0 & 0 & 0 & 1 & - 6\end{array} \right]​1000​0100​00−10​0001​249−6​​

Solve the system using Gaussian elimination or Gauss-Jordan elimination. - 5x−7y=−3−6x+5y=7\begin{aligned}5 x - 7 y & = - 3 \\- 6 x + 5 y & = 7\end{aligned}5x−7y−6x+5y​=−3=7​