I have a few questions I need help with! Please explain if my answer was not right how you got to it. I got 1.) B 2.) A 3.) C
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
A.)(1,1,2)
B.)(0, -2)
C.) (0,-3)
D.) (-1,-2)
2. Solve the system by triangularizing the augmented matrix and using back substitution.
6x=y-4
3y=-13-7x
a.) (-1,-3)
b.) (0,-2)
C.) (0,-3)
D.) (-1,-2)
3. What row operation would produce zero beneath the first entry in the diagonal?
1 2 | -1
8 -2 | -2
A.) 8R1-R2--> R2
B.) -8R1-R2 -->R2
C.) -R1+R2 --> R2
D.) 8R1-R2 --> R2
for 2 answer is <d>
1 is ok
1. To solve the system by triangularizing the augmented matrix and using back substitution, we can begin by writing down the augmented matrix:
| -3 1 -1 | -4 |
| 6 -2 2 | 8 |
|-12 4 -4 |-16 |
Next, we'll perform row operations to transform this matrix into the triangular form. Remember, the goal is to create zeros beneath the diagonal entries. Let's start by eliminating the x-coefficient in the second equation:
| -3 1 -1 | -4 |
| 6 -2 2 | 8 |
|-12 4 -4 |-16 |
Row2 = Row2 + 2 * Row1
Row3 = Row3 - 4 * Row1
| -3 1 -1 | -4 |
| 0 0 0 | 0 |
| 0 8 -8 | 0 |
As you can see, the second row contains all zeros. Therefore, it represents an equation with no information (0 = 0), meaning the system is linearly dependent. We can express the solution in terms of a parameter.
-3x + y - z = -4 => -3x + y - z = -4 => y - z = -4 + 3x => y = 4 - 3x + z
Now, choose a parameter to represent z, let's say z = t. Substitute this value into the equation for y:
y = 4 - 3x + t
For the x-value, we can solve for it by using the third equation (-12x + 4y - 4z = -16):
-12x + 4(4 - 3x + t) - 4t = -16
-12x + 16 - 12x + 4t - 4t = -16
-24x + 16 = -16
-24x = -32
x = 2/3
Finally, substitute the values of x and z into the equation for y to get the full solution:
(x, y, z) = (2/3, 4 - 3(2/3) + t, t)
Simplifying y:
y = 4 - 2 + t
y = 2 + t
So, the solution to the system is (x, y, z) = (2/3, 2 + t, t), where t is a parameter.
Therefore, the correct answer is C.) (0, -3).
2. Let's solve the system by triangularizing the augmented matrix and using back substitution:
| 6 -1 0 | -4 |
| -7 3 1 |-13 |
First, we can perform row operations to eliminate the x-coefficient in the second equation:
Row2 = Row2 + (7/6) * Row1
| 6 -1 0 | -4 |
| 0 4 1 |-10 |
The augmented matrix is now in triangular form. We can proceed with back substitution by solving the second equation for y:
4y + z = -10
4y = -10 - z
y = (-10 - z) / 4
Next, substitute this value for y in the first equation to solve for x:
6x - (-10 - z) / 4 + 0 = -4
6x = -4 + 10 + z/4
6x = 6 + z/4
x = 1 + (z/4) / 6
x = 1 + z/24
x = (24 + z) / 24
The full solution for the system is (x, y) = ((24 + z) / 24, (-10 - z) / 4).
Comparing this with the given answer choices, we can see that the correct answer is D.) (-1, -2).
3. To produce zero beneath the first entry in the diagonal, we need to perform a row operation. Let's examine the given matrix:
| 1 2 | -1 |
| 8 -2 | -2 |
To eliminate the 8 beneath the first entry in the diagonal, we can subtract 8 times the first row from the second row. It can be written as:
Row2 = Row2 - 8 * Row1
Applying this row operation to the matrix:
| 1 2 | -1 |
| 0 -18 | 6 |
As you can see, after the row operation, the entry beneath the first entry in the diagonal becomes zero. Therefore, the correct answer is A.) 8R1 - R2 -> R2.