If $b$ and $c$ are non-zero real numbers,$A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$,then $\det(A+B) = $

  • A
    $3$
  • B
    $1$
  • C
    $-1$
  • D
    $0$

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If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.

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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

$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} = \dots$

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

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