The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

  • A
    -$17$
  • B
    -$16$
  • C
    $16$
  • D
    $17$

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For all values of $A, B, C$ and $P, Q, R$,the value of $\left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$ is

If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$

If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right| = 0$,then the triangle is

$\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = $

The solution$(s)$ of the equation $\left| \begin{matrix} x & a & b \\ a & x & a \\ b & b & x \end{matrix} \right| = 0$ is/are:

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