Mathematics - Vectors and Matrices
46. An arbitary vector X is an eigen vector of the matrix
then (a,b)=
- Option : B
- Explanation : Since matrix is triangular, the eigen values are a, a, b.
(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)
(a, b) = (1, 1)47. For which value of k, the following system is consistent?
2x-5ky+6z=0
kx+2y-2z=0
2x+2y-kz=0
- Option : B
- Explanation :
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
or 2
0
-5k3+ 20k - 4k + 8 + 12k - 24 = 048. The value of λ for which the equations
2x + y + 2z = 0
x + y + 3z = 0
4x + y + λz = 0
have non-zero solution, is
- Option : D
- Explanation : Equivalent matrix equation is
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),
a1x + a2y = 0
b1x + b2y = 0
where a1, a2 , b1, b2 are real numbers,
has a non-trivial solutions if
50. The solution of the given matrix equation is
- Option : A
- Explanation : Let matrices
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
