The system of equations:
is
(1) consistent and independent if
AE ≠ BD
(2) consistent and dependent if
AE = BD and AF = CD
Here A=k, B=3, C=k-2, D=12, E=k, F=k
Since we want them to be inconsistent, we must rule out
both case (1) and case (2).
To rule out case (1) we require AE ≠ BD to be false, so we must
require:
AE = BD
k(k) = 3(12)
kČ = 36
k = ±6
But we also must check to see that k = ±6 also rules out
case (2)
That is we must be sure that AF ≠ CD
AF ≠ CD
k(k) ≠ (k-2)(12)
±6(±6) ≠ (±6-2)(12)
Taking the + we have
36 ≠ (6-2)(12)
36 ≠ (4)(12)
36 ≠ 48
which is true.
Taking the - we have
36 ≠ (-6-2)(12)
36 ≠ (-8)(12)
36 ≠ -96
which is also true.
So case (2) is ruled out be taking k = ±6
It was necessary to rule out case (2), however.
Edwin