$$\text { If } \begin{aligned} f(x) & =\left|\begin{array}{ccc} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{array}\right| \\ & =A+B x+C x^2+\ldots, \text { then } A \text { is equal to ............. .} \end{aligned}$$
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.
If $A$ and $B$ are matrices of order 3 and $|A|=5,|B|=3$, then $|3 A B|=27 \times 5 \times 3=405$.