Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero
Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
Statement $-1$ is false, Statement $-2$ is true
Statement $-1$ is true, Statement $-2$ is false
If $\alpha , \beta \, and \, \gamma$ are real numbers , then $D = \left|{\begin{array}{*{20}{c}}1&{\cos \,(\beta \, - \,\alpha )}&{\cos \,(\gamma \, - \,\alpha )}\\{\cos \,(\alpha \, - \,\beta )}&1&{\cos \,(\gamma \, - \,\beta )}\\{\cos \,(\alpha \, - \,\gamma )}&{\cos \,(\beta \, - \,\gamma )}&1 \end{array}} \right|$ =
$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $
Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :
If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\2&{x + 2}&2\\3&3&{x + 3}\end{array}\,} \right| = 0,$ then $x$ is
If $a,b,c$ and $d $ are complex numbers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}2&{a + b + c + d}&{ab + cd}\\{a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)}\\{ab + cd}&{ab(c + d) + cd(a + d)}&{2abcd}\end{array}} \right|$is