If \((\mathrm{cosec}\,\alpha + 2)\,I = I\), where I is the identity matrix of order 2 and \(0 \lt \alpha \lt 2\pi\), then \(\alpha\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3\pi}{2}\)
The left side is a scalar matrix with diagonal elements \(\mathrm{cosec}\,\alpha + 2\). By equality of matrices, its diagonal elements must equal those of I, which are 1.
\(\mathrm{cosec}\,\alpha + 2 = 1\)
\(\mathrm{cosec}\,\alpha = -1\)
\(\sin\alpha = -1\)
In \(0 \lt \alpha \lt 2\pi\), the only such angle is
\( \displaystyle \alpha = \frac{3\pi}{2} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{2}\) — Incorrect. Here \(\mathrm{cosec}\,\alpha = 1\), so the scalar is \(1 + 2 = 3\), giving 3I, not I.
Option (3) \(\displaystyle\frac{\pi}{6}\) — Incorrect. Here \(\mathrm{cosec}\,\alpha = 2\), so the scalar is 4.
Option (4) \(\displaystyle\frac{7\pi}{6}\) — Incorrect. Here \(\mathrm{cosec}\,\alpha = -2\), so the scalar is 0, giving the zero matrix.