The value of $\left|\begin{array}{ccc}x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x\end{array}\right|$ is
If $A$ is a matrix of order $3 \times 3$, then $|3 A|$ is equal to ............. .
If $A$ is a matrix of order $3 \times 3$, then $|3 A|=3 \times 3 \times 3|A|=27|A|$
If $A$ is invertible matrix of order $3 \times 3$, then $\left|A^{-1}\right|$ is equal to ............. .
If $A$ is invertible matrix of order $3 \times 3$, then $\left|A^{-1}\right|=\frac{1}{|A|}$. $\left[\right.$ since, $\left.|A| \cdot\left|A^{-1}\right|=1\right]$
If $x, y, z \in R$, then the value of $\left|\begin{array}{lll}\left(2^x+2^{-x}\right)^2 & \left(2^x-2^{-x}\right)^2 & 1 \\ \left(3^x+3^{-x}\right)^2 & \left(3^x-3^{-x}\right)^2 & 1 \\ \left(4^x+4^{-x}\right)^2 & \left(4^x-4^{-x}\right)^2 & 1\end{array}\right|$ is
We have,
$$\begin{aligned} & \left|\begin{array}{lll} \left(2^x+2^{-x}\right)^2 & \left(2^x-2^{-x}\right)^2 & 1 \\ \left(3^x+3^{-x}\right)^2 & \left(3^x-3^{-x}\right)^2 & 1 \\ \left(4^x+4^{-x}\right)^2 & \left(4^x-4^{-x}\right)^2 & 1 \end{array}\right| \\ & =\left|\begin{array}{lll} \left(2 \cdot 2^x\right)\left(2 \cdot 2^{-x}\right) & \left(2^x-2^{-x}\right)^2 & 1 \\ \left(2 \cdot 3^x\right)\left(2 \cdot 3^{-x}\right) & \left(3^x-3^{-x}\right)^2 & 1 \\ \left(2 \cdot 4^x\right)\left(2 \cdot 4^{-x}\right) & \left(4^x-4^{-x}\right)^2 & 1 \end{array}\right|\quad \begin{array}{r} {\left[\because(a+b)^2-(a-b)^2=4 a b\right]} \\ {\left[\because C_1 \rightarrow C_1-C_2\right]} \end{array} \end{aligned}$$
$=\left|\begin{array}{lll}4 & \left(2^x-2^{-x}\right)^2 & 1 \\ 4 & \left(3^x-3^{-x}\right)^2 & 1 \\ 4 & \left(4^x-4^{-x}\right)^2 & 1\end{array}\right|=0 \quad$ [since, $C_1$ and $C_3$ are proportional to each other]
If $\cos 2 \theta=0$, then $\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^2$ is equal to ............. .
$$\begin{aligned} &\text { Since, } \cos 2 \theta=0\\ &\begin{array}{lc} \Rightarrow & \cos 2 \theta=\cos \frac{\pi}{2} \Rightarrow 2 \theta=\frac{\pi}{2} \\ \Rightarrow & \theta=\frac{\pi}{4} \\ \therefore & \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ \therefore & \left|\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right|^2 \end{array} \end{aligned}$$
$$\begin{aligned} &\text { Expanding along } R_1 \text {, }\\ &=\left[-\frac{1}{\sqrt{2}}\left(\frac{1}{2}\right)+\frac{1}{\sqrt{2}}\left(-\frac{1}{2}\right)\right]^2=\left[\frac{-2}{2 \sqrt{2}}\right]^2=\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2} \end{aligned}$$