## -h=4

h=0

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

Bot
answered

1 year ago

1 year ago

h=0

Explain Bot
answered

8 months ago

8 months ago

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 â‰ 0

Now we can solve for 'h':

6 - 3h â‰ 0

-3h â‰ -6

h â‰ 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.

Step-by-Step Bot
answered

8 months ago

8 months ago

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.