1
NDA 2017 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
The adjoint of the matrix $$A = \left[ {\matrix{ 1 & 0 & 2 \cr 2 & 1 & 0 \cr 0 & 3 & 1 \cr } } \right]$$ is
A
$$\left[ {\matrix{ { - 1} & 6 & 2 \cr { - 2} & 1 & { - 4} \cr 6 & 3 & 1 \cr } } \right]$$
B
$$\left[ {\matrix{ 1 & 6 & { - 2} \cr { - 2} & 1 & 4 \cr 6 & { - 3} & 1 \cr } } \right]$$
C
$$\left[ {\matrix{ 6 & 1 & 2 \cr 4 & { - 1} & 2 \cr 6 & 3 & { - 1} \cr } } \right]$$
D
$$\left[ {\matrix{ { - 6} & 2 & 1 \cr 4 & { - 2} & 1 \cr 3 & 1 & { - 6} \cr } } \right]$$
2
NDA 2017 Paper 2
MCQ (Single Correct Answer)
+2.5
-0.83
If a, b, c are non-zero real numbers, then the inverse of the matrix $$A = \left[ {\matrix{ a & 0 & 0 \cr 0 & b & 0 \cr 0 & 0 & c \cr } } \right]$$ is equal to
A
$$\left[ {\matrix{ {{a^{ - 1}}} & 0 & 0 \cr 0 & {{b^{ - 1}}} & 0 \cr 0 & 0 & {{c^{ - 1}}} \cr } } \right]$$
B
$${1 \over {abc}}\left[ {\matrix{ {{a^{ - 1}}} & 0 & 0 \cr 0 & {{b^{ - 1}}} & 0 \cr 0 & 0 & {{c^{ - 1}}} \cr } } \right]$$
C
$${1 \over {abc}}\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
D
$${1 \over {abc}}\left[ {\matrix{ a & 0 & 0 \cr 0 & b & 0 \cr 0 & 0 & c \cr } } \right]$$
3
NDA 2017 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
If $$A = \left[ {\matrix{ {\cos \alpha } & {\sin \alpha } \cr { - \sin \alpha } & {\cos \alpha } \cr } } \right]$$, then what is AAT equal to (where, AT is transpose of A)?
A
Null matrix
B
Identity matrix
C
A
D
$$-$$ A
4
NDA 2017 Paper 1
MCQ (Single Correct Answer)
+2.5
-0.83
$$A = \left[ {\matrix{ {x + y} & y \cr x & {x - y} \cr } } \right]$$, $$B = \left[ {\matrix{ 3 \cr { - 2} \cr } } \right]$$ and $$C = \left[ {\matrix{ 4 \cr { - 2} \cr } } \right]$$. If AB = C, then what is A2 equal to?
A
$$\left[ {\matrix{ 4 & 8 \cr { - 4} & { - 16} \cr } } \right]$$
B
$$\left[ {\matrix{ 4 & { - 4} \cr 8 & { - 16} \cr } } \right]$$
C
$$\left[ {\matrix{ { - 4} & { - 8} \cr 4 & {12} \cr } } \right]$$
D
$$\left[ {\matrix{ { - 4} & { - 8} \cr 8 & {12} \cr } } \right]$$
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