Statement-1 : The determinant of a matrix A=[aij] of order 5×5 where aij+aji=0 for i≠j is zero Statement-2 : The determinant of a skew symmetric matrix of odd order is zero.

Statement-1 : The determinant of a matrix A=[aij] of order 5×5 where aij+aji=0 for i≠j is zero Statement-2 : The determinant of a skew symmetric matrix of odd order is zero.

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Statement-1 : The determinant of a matrix $A = [a_{i j}]$ of order $5 \times 5$ where $a_{i j} + a_{j i} = 0 f o r i \neq j$ is zero
Statement-2 : The determinant of a skew symmetric matrix of odd order is zero.

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Statement-1 is True, Statement-2 is True; Statement-2 is a correct a explanation for Statement1.

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

answer is D.

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Detailed Solution

$| \begin{array}{l} \begin{matrix} 1 & 1 & 1 & 6 \end{matrix} \\ \begin{matrix} 1 & 2 & 3 & 14 \end{matrix} \\ \begin{matrix} α & β & 7 & 3 \end{matrix} \end{array} |$ by using $R_{2} - R_{1}, R_{3} - R_{1}$ it become $| \begin{array}{l} \begin{matrix} 1 & 1 & 1 & 6 \end{matrix} \\ \begin{matrix} 0 & 1 & 2 & 8 \end{matrix} \\ \begin{matrix} 0 & β - α & 7 - α & 3 - 6 α \end{matrix} \end{array} |$
Then by using $R_{3} - (β - α) R_{2}$ it become $| \begin{array}{l} \begin{matrix} 1 & 1 & 1 & 6 \end{matrix} \\ \begin{matrix} 0 & 1 & 2 & 8 \end{matrix} \\ \begin{matrix} 0 & 0 & α - 2 β + 7 & 2 α - 8 β + 3 \end{matrix} \end{array} |$