(X1,X2,X3) is an arbitary eigen vector, say corresponding to 1, then
X2 X3 being not zero, we have
X1 = X1 ; aX2 = X2
which gives a = 1
and bX3 = X3
which gives b - 1
(a, b) = (1, 1)
Equations are consistent, if rank of A and that of k are equal. But in this case it is always true. Hence the equations
will have a trivial solution if IAI 0
Therefore only non- trivial solution will exist if IAI = 0
2(-2k+4) + k(-k2+4+6)(2k-4) = 0
-5k3+ 20k - 4k + 8 + 12k - 24 = 0
5k3 - 28k + 16 = 0
5k3 - 10k2 + 10k2 - 20k - 8k + 16 = 0
(5k2+ 10k - 8)(k - 2) = 0
or 2
In order that the given system of equations may have non-zero solution, the rank of A should be less than 3.
This requires that
Interchanging R1 and R2
By (R2 - 2R1) and (R3 - 4R1),
AX = B
Multiplying both sides by A-1
X = A-1B
But as B = 0 therfore X = 0
Hence
x1 = 0, x2 = 0, x3 = 0
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(a, b) = (1, 1)
0
-5k3+ 20k - 4k + 8 + 12k - 24 = 0