If $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ a & b & c \\ l & m & n\end{array}\right|=(-1)^K\left|\begin{array}{ccc}m & n & l \\ b & c & a \\ \beta & \gamma & \alpha\end{array}\right|$, then the least value of $K$ is

  • A
    $2$
  • B
    $3$
  • C
    $4$
  • D
    $5$

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Similar Questions

Let $ \Delta = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} $ and $ \Delta_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix} $,then:

If $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then:

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$\left|\begin{array}{ccc}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b\end{array}\right|$ is equal to

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

Show that $\left|\begin{array}{ccc}a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z\end{array}\right|=0$.

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