NDA 2018 Paper 1 | Matrices Question 31 | Mathematics

What is the inverse of the matrix $$A = \left( {\matrix{ {\cos \theta } & {\sin \theta } & 0 \cr { - \sin \theta } & {\cos \theta } & 0 \cr 0 & 0 & 1 \cr } } \right)$$ ?

$$\left( {\matrix{ {\cos \theta } & { - \sin \theta } & 0 \cr {\sin \theta } & {\cos \theta } & 0 \cr 0 & 0 & 1 \cr } } \right)$$

$$\left( {\matrix{ {\cos \theta } & 0 & {\sin \theta } \cr 0 & 1 & 0 \cr {\sin \theta } & 0 & {\cos \theta } \cr } } \right)$$

$$\left( {\matrix{ 1 & 0 & 0 \cr 0 & {\cos \theta } & { - \sin \theta } \cr 0 & {\sin \theta } & {\cos \theta } \cr } } \right)$$

$$\left( {\matrix{ {\cos \theta } & {\sin \theta } & 0 \cr { - \sin \theta } & {\cos \theta } & 0 \cr 0 & 0 & 1 \cr } } \right)$$

NDA 2019 Paper 2

MCQ (Single Correct Answer)

+2.5

-0.83

If $$A = \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$$, then the expression A³ $$-$$ 2A² is

a null matrix

an identity matrix

equal to A

equal to $$-$$ A

NDA 2019 Paper 2

MCQ (Single Correct Answer)

+2.5

-0.83

If $$A = \left( {\matrix{ 1 & 2 \cr 2 & 3 \cr 3 & 4 \cr } } \right)$$ and $$B = \left( {\matrix{ 1 & 2 \cr 2 & 1 \cr } } \right)$$, then which one of the following is correct?

Both AB and BA exist

Neither AB nor BA exists

AB exists but BA does not exist

AB does not exist but BA exists

NDA 2019 Paper 1

MCQ (Single Correct Answer)

+2.5

-0.83

If $$B = \left[ {\matrix{ 3 & 2 & 0 \cr 2 & 4 & 0 \cr 1 & 1 & 0 \cr } } \right]$$, then what is adjoint of B equal to?

$$\left[ {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr { - 2} & { - 1} & 8 \cr } } \right]$$

$$\left[ {\matrix{ 0 & 0 & { - 2} \cr 0 & 0 & { - 1} \cr 0 & 0 & 8 \cr } } \right]$$

$$\left[ {\matrix{ 0 & 0 & 2 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right]$$