-6x + 5y + 6z = -5
3x + 2y + 6z = -5
-6x + 23y + hz = k
For this to have infinitely many solutions, the coefficient determinant must
equal 0:
We evaluate the determinant by copying over the first two columns on the
right
Add the products of the three upper left to lower right diagonal elements:
(-6)(2)(h)+(5)(6)(-6)+(6)(3)(23)
-12h-180+414
-12h+234
Now subtract the products of the three upper right to lower left diagonal elements:
-(6)(2)(-6)-(-6)(6)(23)-(5)(3)(h)
72+828-15h
900-15h
We combine those and get -12h+234+900-15h = -27h+1134
Set that equal to 0
-27h+1134 = 0
-27h = -1134
h = 42
Now our system becomes:
-6x + 5y + 6z = -5
3x + 2y + 6z = -5
-6x + 23y + 42z = k
Eliminate z from the first two by multiplying the
1st equation by -1 and adding. That gives:
9x - 3y = 0
Now eliminate z from the second and third by
multiplying the second by -7 and adding.
-21x - 14y - 42z = 35
-6x + 23y + 42z = k
---------------------------
-27x + 9y = 35+k
Now we multiply the equation 9x - 3y = 0 through
by 3 and add it to that result:
-27x + 9y = 35+k
27x - 9y = 0
-----------------------------
0 = 35+k
35 = k
So h = 42 and k = 35
That's the answer. But eventually, you'll have to give the solution
in terms of z. When you do, the solution will be:
The solution is (x,y,z) =
Edwin
