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JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Matrices

If

A = (0 sin α sin α 0)

and

det (A^{2} - \frac{1}{2} I) = 0

, then a possible value of

α

is :

(A) $\frac{π}{6}$
(B) $\frac{π}{3}$
(C) $\frac{π}{4}$ ✓ Correct
(D) $\frac{π}{2}$

Answer: (C)

Solution

First compute

A^{2} = (0 sin α sin α 0) (0 sin α sin α 0) = (sin^{2} α 0 0 sin^{2} α) = sin^{2} α \cdot I

. Then

A^{2} - \frac{1}{2} I = (sin^{2} α - \frac{1}{2}) I

. The determinant is

(sin^{2} α - \frac{1}{2})^{2} = 0

, so

sin^{2} α = \frac{1}{2}

, giving

sin α = \pm \frac{1}{2}

. Thus

α = \frac{π}{4}

(or

\frac{3 π}{4}

, etc.).

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option

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