$\left|\begin{array}{ccc}0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|$ is equal to ............ .
We have, $\left|\begin{array}{ccc}0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|=\left|\begin{array}{ccc}z-x & x y z & x-z \\ z-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|\quad$ $$ \left[\because C_1 \rightarrow C_1-C_3\right]$$
$$ =(z-x)\left|\begin{array}{ccc} 1 & x y z & x-z \\ 1 & 0 & y-z \\ 1 & z-y & 0 \end{array}\right|\quad$$ [taking $(z-x)$ common from column 1]
$$\begin{aligned} &\text { Expanding along } R_1 \text {, }\\ &\begin{aligned} & =(z-x)[1 \cdot\{-(y-z)(z-y)\}-x y z(z-y)+(x-z)(z-y)] \\ & =(z-x)(z-y)(-y+z-x y z+x-z) \\ & =(z-x)(z-y)(x-y-x y z) \\ & =(z-x)(y-z)(y-x+x y z) \end{aligned} \end{aligned}$$
$$\text { If } \begin{aligned} f(x) & =\left|\begin{array}{ccc} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{array}\right| \\ & =A+B x+C x^2+\ldots, \text { then } A \text { is equal to ............. .} \end{aligned}$$
Since,
$$\begin{aligned} &f(x)=(1+x)^{17}(1+x)^{23}(1+x)^{41}\left|\begin{array}{ccc} 1 & (1+x)^2 & (1+x)^6 \\ 1 & (1+x)^6 & (1+x)^{11} \\ 1 & (1+x)^2 & (1+x)^6 \end{array}\right|=0\\ &\text { [since, } R_1 \text { and } R_3 \text { are identical] }\\ \therefore\quad &A=0 \end{aligned}$$
$\left(A^3\right)^{-1}=\left(A^{-1}\right)^3$, where $A$ is a square matrix and $|A| \neq 0$.
$(a A)^{-1}=\frac{1}{a} A^{-1}$, where $a$ is any real number and $A$ is a square matrix.
$\left|A^{-1}\right| \neq|A|^{-1}$, where $A$ is a non-singular matrix.