If $D = \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix}$,then $D =$

  • A
    $1 + a^2 + b^2 + c^2$
  • B
    $a^2 + b^2 + c^2$
  • C
    $(a + b + c)^2$
  • D
    None of these

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If $\omega$ is an imaginary cube root of unity, then the value of the determinant $\left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$ is

Statement-$1$: The determinant of a skew-symmetric matrix of order $3$ is zero.
Statement-$2$: For any square matrix $A$ of order $n$,$\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$.

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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$\left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \\ 1 & ca+bd & c^2a^2+b^2d^2 \\ 1 & ab+cd & a^2b^2+c^2d^2\end{array}\right|=$

Evaluate $\Delta=\left|\begin{array}{lll}3 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 2 & 3\end{array}\right|$

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