1
NDA 2017 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
The system of equations kx + y + z = 1, x + ky + z = k and x + y + kz = k2 has no solution if k equals.
A
0
B
1
C
$$-$$ 1
D
$$-$$ 2
2
NDA 2017 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
The value of the determinant $$\left| {\matrix{ {1 - \alpha } & {\alpha - {\alpha ^2}} & {{\alpha ^2}} \cr {1 - \beta } & {\beta - {\beta ^2}} & {{\beta ^2}} \cr {1 - \gamma } & {\gamma - {\gamma ^2}} & {{\gamma ^2}} \cr } } \right|$$ is equal to
A
($$\alpha$$ $$-$$ $$\beta$$) ($$\beta$$ $$-$$ $$\gamma$$) ($$\alpha$$ $$-$$ $$\gamma$$)
B
($$\alpha$$ $$-$$ $$\beta$$) ($$\beta$$ $$-$$ $$\gamma$$) ($$\gamma$$ $$-$$ $$\alpha$$)
C
($$\alpha$$ $$-$$ $$\beta$$) ($$\beta$$ $$-$$ $$\gamma$$) ($$\gamma$$ $$-$$ $$\alpha$$) ($$\alpha$$ + $$\beta$$ + $$\gamma$$)
D
0
3
NDA 2017 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
If p + q + r = a + b + c = 0, then the determinant $$\left| {\matrix{ {pa} & {qb} & {rc} \cr {qc} & {ra} & {pb} \cr {rb} & {pc} & {qa} \cr } } \right|$$ equals
A
0
B
1
C
pa + qb + rc
D
pa + qb + rc + a + b + c
4
NDA 2017 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
If $$A = \left[ {\matrix{ \alpha & 2 \cr 2 & \alpha \cr } } \right]$$ and det (A3) = 125, then $$\alpha$$ is equal to
A
$$\pm$$ 1
B
$$\pm$$ 2
C
$$\pm$$ 3
D
$$\pm$$ 5
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