Statement-1: The determinant of a skew-symmet | vedclass

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(A) For Statement-$1$: $A$ skew-symmetric matrix $A$ satisfies $A^T = -A$. Taking the determinant on both sides,we get $\det(A^T) = \det(-A)$. Since $

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Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

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(A) For Statement-$1$: $A$ skew-symmetric matrix $A$ satisfies $A^T = -A$. Taking the determinant on both sides,we get $\det(A^T) = \det(-A)$. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$ for a matrix of order $n$,we have $\det(A) = (-1)^n \det(A)$. For $n=3$,$\det(A) = -\det(A)$,which implies $2 \det(A) = 0$,so $\det(A) = 0$. Thus,Statement-$1$ is true.For Statement-$2$: The property $\det(A^T) = \det(A)$ is always true. The property $\det(-A) = (-1)^n \det(A)$ is also true for a matrix of order $n$. Therefore,Statement-$2$ is true.Since Statement-$2$ provides the mathematical properties used to prove Statement-$1$,it is the correct explanation for Statement-$1$.

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 $\det(-A) = (-1)^n \det(A)$.

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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 $\det(-A) = (-1)^n \det(A)$.