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For the following system to be consistent,

6x + 4y - 5z = 7
7x + 2y + kz = -5
32x + 16y - z = 6
we must have, k ≠ ?

1 Answer

5 votes

Final answer:

The system will be consistent for any value of k.

Step-by-step explanation:

The given system of equations is:

6x + 4y - 5z = 7

7x + 2y + kz = -5

32x + 16y - z = 6

To check if the system is consistent, we can eliminate variables x and y by multiplying the first equation by 2 and the second equation by 3. This will give us:

12x + 8y - 10z = 14

21x + 6y + 3kz = -15

Subtracting the first equation from the second equation:

9x - 14z = -29

From the third equation, we can write:

32x + 16y = z + 6

Substituting the value of z from the second equation into the first equation:

9x - 14(32x + 16y - 6) = -29

Simplifying:

9x - 448x - 224y + 84 = -29

-439x - 224y = -113

Now we have a system of two equations:

9x - 14z = -29

-439x - 224y = -113

If the system is consistent, there will be a unique solution for x and y. To check if the system has a unique solution, we can calculate the determinant:

D = (9)(-224) - (-14)(-439)

D = 1996

Since the determinant is not zero, the system has a unique solution for x and y. Therefore, for the system to be consistent, k can be any value.

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