If $\Delta = \begin{vmatrix} x & y & z \\ p & q & r \\ a & b & c \end{vmatrix}$,then $\begin{vmatrix} x & 2y & z \\ 2p & 4q & 2r \\ a & 2b & c \end{vmatrix}$ equals

  • A
    $\Delta^2$
  • B
    $4\Delta$
  • C
    $3\Delta$
  • D
    None of these

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Similar Questions

Using the property of determinants and without expanding,prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$.

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|=a^{3}$

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

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If $\left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array} \right| = \alpha (a+b)(b+c)(c+a) \neq 0$,then $\alpha$ is equal to

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

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