I have a few questions I need help with!

1. Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
-3x+y-z=-4
6x-2y+2z=8
-12x+4y-4z=-16

2. Solve the system by triangularizing the augmented matrix and using back substitution.
6x=y-4
3y=-13-7x

3. What row operation would produce zero beneath the first entry in the diagonal?
1 2 | -1
8 -2 | -2

1. Ah, linear algebra, keeping mathematicians around since forever. Let's dive into it!

We start by writing the system of equations as an augmented matrix:

| -3 1 -1 | -4 |
| 6 -2 2 | 8 |
|-12 4 -4 |-16 |

Now we want to triangularize the matrix, meaning we want to have zeros below the diagonal elements. We can do that by performing row operations. Let's start:

R2 = R2 + 2R1
R3 = R3 + 4R1

| -3 1 -1 | -4 |
| 0 0 0 | 0 |
| 0 8 -8 | -8 |

Great! Now, we can see that the second row is all zeros, which means that the system is linearly dependent. This basically means that one of the equations is a linear combination of the others. In this case, we can see that the second equation is just twice the first equation, so we can say y = 2x.

Now, we substitute that into the third row:

0 + 8y - 8z = -8

Substituting y = 2x:

0 + 8(2x) - 8z = -8
16x - 8z = -8
8x - 4z = -4
2x - z = -1

And there you have it! The solution, when expressed in terms of a parameter, is:
x = s (where s is any real number),
y = 2s,
z = 2s + 1.

2. Let's continue our linear algebra adventure!

The augmented matrix for this system is:

| 6 -1 0 | 4 |
| -7 3 1 |-13 |

To triangularize it, we can perform the following row operation:

R2 = R2 + (7/6)R1

| 6 -1 0 | 4 |
| 0 2/3 1 | -13+7/2 |

| 6 -1 0 | 4 |
| 0 2/3 1 | -19/2|

Perfect! Now let's apply back substitution:

2/3 * y + z = -19/2
2y + 3z = -19/3

Now let's solve for x:

6x - y = 4
6x = y + 4
x = (y + 4)/6

And there you have it! The solution to the system is:
x = (y + 4)/6,
y = y (yep, just y),
z = -19/2 - (2/3)y.

3. Ah, a row operation riddle! Let me put on my thinking cap.

To produce a zero beneath the first entry in the diagonal, we need to perform the following row operation:

R2 = R2 - 8R1

| 1 2 | -1 |
| 0 -18 | 6 |

And voila! We now have a zero beneath the first entry in the diagonal. It's like magic, but with more numbers!

Sure, I can help you with that! Let's go through each question step by step:

1. To solve the system, we first write the augmented matrix:

| -3 1 -1 | -4 |
| 6 -2 2 | 8 |
| -12 4 -4 | -16 |

Next, we'll perform row operations to triangularize the matrix:

R2 = R2 + 2R1
R3 = R3 - 4R1

| -3 1 -1 | -4 |
| 0 0 0 | 0 |
| 0 0 0 | 0 |

Now, we can see that the system is linearly dependent since the second and third row are all zeros. To express the solution in terms of a parameter, let's assign a value to z, let's say z = t. Then we can express x and y in terms of t.

From the first row, we have:

-3x + y - (-t) = -4
-3x + y + t = -4
y = 3x - 4 - t

So the solution can be written as:

x = t
y = 3t - 4
z = t

2. Let's start by writing the augmented matrix:

| 6 -1 0 | -4 |
| 7x 3 -1 | -13 |

Next, let's perform row operations to triangularize the matrix:

R2 = R2 - (7/6)R1

| 6 -1 0 | -4 |
| 0 4 -1 | -1/6 |

Now we can use back substitution to solve for x and y:

From the second row, we have:

4y - z = -1/6
z = 4y + 1/6

From the first row, we have:

6x - y = -4
6x = y - 4
x = (1/6)y - 2/3

So the solution can be written as:

x = (1/6)y - 2/3
y = y (which is a free variable since it does not depend on other variables)
z = 4y + 1/6

3. To produce a zero beneath the first entry in the diagonal, we can perform the following row operation:

R2 = R2 - 8R1

The resulting matrix will be:

| 1 2 | -1 |
| 0 -18 | 6 |

Now, beneath the first entry in the diagonal, we have a zero.

Sure! Let's go through each question step by step.

1. Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
To solve the system using triangularization and back substitution, we need to convert the system of equations into an augmented matrix and perform row operations.

First, let's write the augmented matrix:

| -3 1 -1 -4 |
| 6 -2 2 8 |
|-12 4 -4 -16 |

We can start by using row operations to transform the matrix into a lower triangular form.

Step 1: Multiply the first row by (1/(-3)) to make the leading coefficient of the first row equal to 1.
|-1 1/3 -1/3 4/3 |
| 6 -2 2 8 |
|-12 4 -4 -16 |

Step 2: Add 6 times the first row to the second row to eliminate the x variable in the second row.
|-1 1/3 -1/3 4/3 |
| 0 0 0 20 |
|-12 4 -4 -16 |

Step 3: Add 12 times the first row to the third row to eliminate the x variable in the third row.
|-1 1/3 -1/3 4/3 |
| 0 0 0 20 |
| 0 12/3 -12/3 -4/3 |

Now, let's rewrite the augmented matrix as a system of equations using this triangularized form:

-x + (1/3)y - (1/3)z = 4/3
0x + 0y + 0z = 20
0x + 4y - 4z = -4/3

Since the second row consists of all zeros, this means that the system is inconsistent, and there is no solution. Therefore, we cannot apply back substitution.

2. Solve the system by triangularizing the augmented matrix and using back substitution.
Let's write the augmented matrix for this system:

| 6 -1 0 -4 |
| -7x + 3y + 0z = -13 |

Step 1: Divide the first row by 6 to make the leading coefficient of the first row equal to 1.
| 1 -1/6 0 -2/3 |
| -7x + 3y + 0z = -13 |

Step 2: Eliminate x from the second row by adding (7/6) times the first row to the second row.
| 1 -1/6 0 -2/3 |
| 0 7/6 0 -20/3 |

Now, let's rewrite the augmented matrix as a system of equations using this triangularized form:

x - (1/6)y + 0z = -2/3
0x + (7/6)y + 0z = -20/3

We can solve using back substitution:

From the second row, we have (7/6)y = -20/3. Solving for y, we get y = -40/7.

Substitute the value of y back into the first row: x - (1/6)(-40/7) = -2/3. Simplifying this equation gives us x = -10/7.

Therefore, the solution to the system is x = -10/7 and y = -40/7.

3. What row operation would produce zero beneath the first entry in the diagonal?
To produce zero beneath the first entry in the diagonal, we need to use a row operation called "row scaling" or "scalar multiplication."

In the given matrix:

| 1 2 | -1 |
| 8 -2 | -2 |

We want to make the second entry (8) in the second row become zero.

We can achieve this by multiplying the first row (-1 2) by 8 and adding it to the second row (8 - 2 -2):

Row 2 <- Row 2 + 8 * Row 1

After performing this row operation, the resulting matrix will be:

| 1 2 | -1 |
| 0 -14 | 6 |

Now, there is a zero beneath the first entry in the diagonal.