WB JEE 2017 | Matrices and Determinants Question 61 | Mathematics

WB JEE 2017

MCQ (Single Correct Answer)

-0.25

The value of det A, where $$A\, = \left( {\matrix{ 1 & {\cos \theta } & 0 \cr { - \cos \theta } & 1 & {\cos \theta } \cr { - 1} & { - \cos \theta } & 1 \cr } } \right)$$, lies

in the closed interval [1, 2]

in the closed interval [0, 1]

in the open interval (0, 1)

in the open interval (1, 2)

WB JEE 2017

MCQ (Single Correct Answer)

-0.5

Let $$A = \left( {\matrix{ 1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr } } \right)$$. Then, for positive integer n, Aⁿ is

$$\left( {\matrix{ 1 & n & {{n^2}} \cr 0 & {{n^2}} & n \cr 0 & 0 & n \cr } } \right)$$

$$\left( {\matrix{ 1 & 1 & {n\left( {{{n + 1} \over 2}} \right)} \cr 0 & 1 & n \cr 0 & 0 & 1 \cr } } \right)$$

$$\left( {\matrix{ 1 & {{n^2}} & n \cr 0 & n & {{n^2}} \cr 0 & 0 & {{n^2}} \cr } } \right)$$

$$\left( {\matrix{ 1 & n & {2n - 1} \cr 0 & {{{n + 1} \over 2}} & {{n^2}} \cr 0 & 0 & {{{n + 1} \over 2}} \cr } } \right)$$

WB JEE 2017

MCQ (Single Correct Answer)

-0.5

Let a, b, c be such that b(a + c) $$ \ne $$ 0. If $$\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {b + 1} & {c - 1} \cr {a - 1} & {b - 1} & {c + 1} \cr {{{( - 1)}^{n + 2}}a} & {{{( - 1)}^{n + 1}}b} & {{{( - 1)}^n}c} \cr } } \right| = 0$$, then the value of n is

any integer

zero

any even integer

any odd integer

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