I need to solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the soloution in terms of a parameter.
-3x+y-z=-4
6x-2y+2z=8
-12x+4y-4z=-16
To solve the system using triangularization and back substitution, follow these steps:
Step 1: Write the augmented matrix of the system:
Coefficients | Constants
-3 1 -1 | -4
6 -2 2 | 8
-12 4 -4 | -16
Step 2: Perform row operations to eliminate variables below the main diagonal:
- Multiply the first row by 2 and add it to the second row
- Multiply the first row by 4 and add it to the third row
New augmented matrix:
-3 1 -1 | -4
0 -2 0 | 0
0 8 -8 | -0
Step 3: Solve for the first variable by using back substitution:
From the second row, we can see that -2y = 0, so y = 0.
Substitute y = 0 into the first row:
-3x + 0 - z = -4
-3x - z = -4
Rearrange the equation to solve for z:
z = -4 + 3x
Now, we have the solution for z in terms of x.
Step 4: Substitute the value of y = 0 and z = -4 + 3x into the third row:
0 + 8(0) - 8(-4 + 3x) = 0
Simplify the equation:
0 + 0 + 32 - 24x = 0
-24x = -32
x = 32/24
x = 4/3
Now we have the value of x.
Step 5: Substitute the values of x = 4/3 and y = 0 into the first row:
-3(4/3) + 0 - z = -4
-4 + z = -4
z = 0
Now we have the value of z.
Step 6: Write the solution:
x = 4/3
y = 0
z = 0
Since the system is linearly independent (no parameter), the solution does not involve any parameters.