The value of the determinant left| begin{arra | vedclass

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(C) Let $\Delta = \left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \ a^3 - b^3 & 0 & c^3 - b^3 \ a^3 - c^3 & b^3 - c^3 & 0 \end{array} ight|$.T

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $\Delta = \left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \ a^3 - b^3 & 0 & c^3 - b^3 \ a^3 - c^3 & b^3 - c^3 & 0 \end{array} ight|$.This is a skew-symmetric determinant of odd order $(3 	imes 3)$.For any skew-symmetric matrix $A$ of odd order $n$,the determinant is given by $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$.Since $n = 3$ (odd),$\det(A) = -\det(A)$,which implies $2 \det(A) = 0$,so $\det(A) = 0$.Alternatively,by observing the rows: $R_1 + R_2 + R_3 = [0, 0, 0]$,which means the rows are linearly dependent.Therefore,the value of the determinant is $0$.

The value of the determinant $\left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \\ a^3 - b^3 & 0 & c^3 - b^3 \\ a^3 - c^3 & b^3 - c^3 & 0 \end{array} \right|$ is equal to:

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