The number of distinct real roots
of $$\left| {\matrix{
{\sin x} & {\cos x} & {\cos x} \cr
{\cos x} & {\sin x} & {\cos x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$ in the interval $- {\pi \over 4} \le x \le {\pi \over 4}$ is :
Solution
$$\left| {\matrix{
{\sin x} & {\cos x} & {\cos x} \cr
{\cos x} & {\sin x} & {\cos x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$$<br><br>Apply : ${R_1} \to {R_1} - {R_2}$ & ${R_2} \to {R_2} - {R_3}$<br><br>$\Rightarrow$ $$\left| {\matrix{
{\sin x - \cos x} & {\cos x - \sin x} & 0 \cr
0 & {\sin x - \cos x} & {\cos x - \sin x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$<br><br>$\Rightarrow$ $${(\sin x - \cos x)^2}\left| {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$<br><br>$\Rightarrow$ ${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$<br><br>$\therefore$ $x = {\pi \over 4}$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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