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  • explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 24 maths

explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 24 maths

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Answer: \frac{1}{2}

Hint: Here, we use basic concept of determinant of matrix.

Given: \cos 2 \theta=0

            \left[\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right]^{2}=?

Solution:

                \cos 2 \theta=0

                \begin{aligned} &\cos 2 \theta=\cos 90^{\circ} \quad\left(\cos 90^{\circ}=0\right) \\ &2 \theta=90^{\circ} \\ &\theta=45^{\circ} \end{aligned}

Let’s put \theta=45^{\circ}  in matrix

                \left[\begin{array}{ccc} 0 & \cos 45^{\circ} & \sin 45^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ} & 0 \\ \sin 45^{\circ} & 0 & \cos 45^{\circ} \end{array}\right]^{2}

                \left[\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right]^{2} \quad\left[\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]

Let’s take \frac{1}{\sqrt{2}} in common

                =\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]

                \begin{aligned} &=\left[\frac{1}{2 \sqrt{2}}[-1 \times(1-0)+1(0-1)]\right]^{2} \\ &=\frac{1}{8} \times[-1-1]^{2} \\ &=\frac{1}{8} \times(-2)^{2} \\ &=\frac{1}{2} \end{aligned}

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