(31) If matrix $$ A $$ is skew symmetric of order $$ 2 imes 2 $$ and $$ |A|=1 $$, then $$ \begin{array}{llll} ext { (a) } A=-A & ext { (b) } A=-1 & ext { (c) } A=A^{-1} & ext { (d) } A^{-1}=-A\end{array} $$

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(31) If matrix $$ A $$ is skew symmetric of order $$ 2 	imes 2 $$ and $$ |A|=1 $$, then $$ \begin{array}{llll}	ext { (a) } A=-A & 	ext { (b) } A=-1 & 	ext { (c) } A=A^{-1} & 	ext { (d) } A^{-1}=-A\end{array} $$

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Let 
$$
A=\begin{pmatrix} 0 & a \ -a & 0 \end{pmatrix}
$$
be a $$2 \times 2$$ skew-symmetric matrix. Its determinant is given by
$$
|A|=(0)(0)-\left(a(-a)\right)=a^2.
$$
Since $$|A|=1$$, we have
$$
a^2=1 \quad \Longrightarrow \quad a=1 \quad \text{or} \quad a=-1.
$$
Thus, the matrix $$A$$ can be either
$$
\begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}\quad \text{or} \quad \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}.
$$

Now, we calculate $$A^{-1}$$. For a $$2 \times 2$$ matrix,
$$
A^{-1}=\frac{1}{|A|}\begin{pmatrix} 0 & -a \ a & 0 \end{pmatrix} =\begin{pmatrix} 0 & -a \ a & 0 \end{pmatrix},
$$
since $$|A|=1$$.

Observe that
$$
-A=\begin{pmatrix} 0 & -a \ a & 0 \end{pmatrix}.
$$
Therefore,
$$
A^{-1}=-A.
$$

Analyzing the given options:

- (a) $$A=-A$$: This would imply $$2A=0$$, i.e. $$A$$ is the zero matrix, which is not the case.
- (b) $$A=-1$$: This equates a matrix with a scalar, which is not meaningful.
- (c) $$A=A^{-1}$$: Since $$A^{-1}=-A$$, this would require $$A=-A$$, implying $$A$$ is the zero matrix.
- (d) $$A^{-1}=-A$$: This is exactly what we have obtained.

Thus, the correct answer is:

$$
\text{(d) } A^{-1}=-A.
$$
The correct answer is (d) $$ A^{-1} = -A $$.

(31) If matrix is skew symmetric of order and , then

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