Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$,where $b_{ij} = 2^{i+j} a_{ij}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$,then the determinant of the matrix $Q$ is:

  • A
    $2^{10}$
  • B
    $2^{11}$
  • C
    $2^{12}$
  • D
    $2^{13}$

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If $x, y, z$ are distinct and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0,$ then show that $1+x y z=0$.

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By using properties of determinants,show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

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$\left| {\begin{array}{ccc} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right| = K \left| {\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right|$,then $K = $

If $a \neq b \neq c$, $\Delta_1=\left|\begin{array}{lll}1 & a^2 & b c \\ 1 & b^2 & c a \\ 1 & c^2 & a b\end{array}\right|$, $\Delta_2=\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ and $\frac{\Delta_1}{\Delta_2}=\frac{6}{11}$, then $11(a+b+c)=$

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

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