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Question: The value of an odd order skew-symmetric determinant is A.Perfect square B.Negative C.\[\pm 1\...

The value of an odd order skew-symmetric determinant is
A.Perfect square
B.Negative
C.±1\pm 1
D.0

Explanation

Solution

Hint: Assume an odd skew-symmetric matrix A. We know that the determinant value of skew-symmetric matrix A and the determinant value of transpose of skew-symmetric matrix A are equal to each other. That is, det(A)=det(AT)\det (A)=det({{A}^{T}}) . Also, the transpose matrix A is equal to the negative of the matrix A. If two matrices are equal then their determinant values are also equal. So, det(AT)=det(A)\det ({{A}^{T}})=det(-A) . We know the formula that, det(A)=(1)ndet(A)det(-A)={{(-1)}^{n}}\det (A) where n is the order of the matrix A. Using this formula for det(A)\det (-A) and take n as odd. If n is odd then (1)n{{(-1)}^{n}} is equal to -1. Now, solve it further.

Complete step-by-step answer:
According to the question, it is given that we have an odd order skew-symmetric determinant.
An odd order means an odd number of rows and columns. It means we have an odd number of rows and columns in an odd skew-symmetric matrix.
Let us assume an odd skew symmetric square matrix having order n, where n is odd.
We know the property that,
det(A)=det(AT)\det (A)=det({{A}^{T}}) ………………………..(1)
where, AT{{\text{A}}^{\text{T}}} is the transpose of matrix A.
We also know that for a skew-symmetric matrix, AT=A{{\operatorname{A}}^{T}}=-A .
If two matrices are equal then its determinant values are also equal, so det(AT)=det(A)\det ({{A}^{T}})=det(-A) …………..(2)
We know the formula that, det(A)=(1)ndet(A)det(-A)={{(-1)}^{n}}\det (A) where n is the order of the matrix A.
From equation (2) and the above equation, we get
det(AT)=(1)ndet(A)\det ({{A}^{T}})={{(-1)}^{n}}\det (A) …………………(3)
Here, n is the order of the matrix A and n is odd.
(1)n=1{{(-1)}^{n}}=-1 ………………….(4)
From equation (3) and equation (4), we have
det(AT)=(1)ndet(A)\det ({{A}^{T}})={{(-1)}^{n}}\det (A)
det(AT)=det(A)\Rightarrow \det ({{A}^{T}})=-\det (A) ……………..(5)
From equation (1) and equation (5), we get
det(A)=det(AT)\det (A)=det({{A}^{T}})

& \Rightarrow \det (A)=-\det (A) \\\ & \Rightarrow 2det(A)=0 \\\ & \Rightarrow det(A)=0 \\\ \end{aligned}$$ So, the determinant value of A is 0. Hence, the option (D) is the correct one. Note: In this question, we have to be careful about the step by step process to solve this question and also one might get confused because the value of n for the formula $$det(-A)={{(-1)}^{n}}\det (A)$$ is not mentioned in the question. But, we don’t need the value of n here because for every odd value of n $${{(-1)}^{n}}$$ is equal to $$-1$$ .