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!