Perform row operations on the augmented matrix as far as necessary to determine whether the system is independent, dependent, or inconsistent. DO ALL WORK BY HAND.
x + 2y + 4z = 6
y + z = 1
x + 3y + 5z =10
If one subtracts the first equation from the last equation...the result is
y+z=4. That is inconsistent with
y+z=1
find the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.
1x+hy=4
3x+6y=8
4
-h=4
h=0
To determine the value(s) of 'h' for which the matrix is the augmented matrix of a consistent linear system, we need to perform row operations until we reach the row echelon form of the matrix.
The given matrix is:
[1, h, 4]
[3, 6, 8]
To perform row operations, we usually start with the first row and transform the matrix until we obtain the row echelon form. Here's how we can do it:
1. If necessary, switch rows to bring a non-zero entry to the top row. Since the first entry in the first row is already non-zero, we can move to the next step.
2. Use row operations to eliminate the non-zero entries below the first entry in the first row. We want to create zeros below the first row.
The goal is to make the matrix of the following form:
[1, h, 4]
[0, ?, ?]
To achieve this, we can subtract 3 times the first row from the second row to eliminate the 3 in the second row:
[1, h, 4]
[0, 6 - 3h, -4]
Now we have:
[1, h, 4]
[0, 6 - 3h, -4]
Since we want to find the value(s) of 'h' for which the matrix is the augmented matrix of a consistent linear system, the second row (the coefficient of y) should not contain any zero values.
So we set 6 - 3h not equal to zero:
6 - 3h ≠ 0
Now we can solve for 'h':
6 - 3h ≠ 0
-3h ≠ -6
h ≠ 2
Therefore, for all values of 'h' except 2, the matrix is the augmented matrix of a consistent linear system.
If 'h' equals 2, then the second row becomes:
[0, 6 - 3(2), -4]
[0, 0, -4]
In this case, the second row consists of zeros, which means the linear system is inconsistent.
So, the value(s) of 'h' for which the matrix is the augmented matrix of a consistent linear system is all values of 'h' except 2.
To find the value(s) of h such that the matrix is the augmented matrix of a consistent linear system, we can perform row operations on the augmented matrix:
1x + hy = 4
3x + 6y = 8
First, let's multiply the first equation by 3 and the second equation by 1:
3(1x + hy) = 3(4)
3x + 3hy = 12
1(3x + 6y) = 1(8)
3x + 6y = 8
Now, let's subtract the first equation from the second equation:
(3x + 6y) - (3x + 3hy) = 8 - 12
3x + 6y - 3x - 3hy = -4
Simplifying, we get:
6y - 3hy = -4
Factoring out y, we have:
y(6 - 3h) = -4
For a consistent linear system, the determinant of the coefficients of the variables (6 - 3h) should be non-zero. Thus, we can set the determinant equal to zero and solve for h:
6 - 3h = 0
Adding 3h to both sides:
6 = 3h
Dividing both sides by 3:
2 = h
Therefore, the value(s) of h such that the matrix is the augmented matrix of a consistent linear system is h = 2.