We are given that,

A =\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) is a skew-symmetric matrix.

We need to find the value of x. 

Let us understand what skew-symmetric matrix is. 

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⇔ A^T = - A 

First, let us find –A.

 - A =-1 x\( \begin{bmatrix}0&1 & -2 \\[0.3em]-1 & 0 &3 \\[0.3em]x & -3 &0\end{bmatrix}\) 

⇒ - A =\( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\) 

Let us find the transpose of A. 

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original. 

In matrix A,

1^st row of A = (0 ,1 ,-2) 

2^nd row of A = (-1 ,0 ,3) 

3^rd row of A = (x ,-3,0) 

In the formation of matrix A^T,

1^st column of A^T = 1^st row of A = (0 ,1 ,-2) 

2^nd column of A^T = 2^nd row of A = (-1,0, 3) 

3^rd column of A^T = 3^rd row of A = (x, -3 ,0)

So,

A^T =\( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\)

Substituting the matrices –A and A^T , we get

- A = A^T

⇒ \( \begin{bmatrix}0&-1 & 2 \\[0.3em]1 & 0 &-3 \\[0.3em]-x & 3 &0\end{bmatrix}\)⁼ \( \begin{bmatrix}0&-1 & x \\[0.3em]1 & 0 &-3 \\[0.3em]-2 & 3 &0\end{bmatrix}\)

We know by the property of matrices,

 \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies, 

a₁₁ = b₁₁, 

a₁₂ = b₁₂, 

a₂₁ = b₂₁ and 

a₂₂ = b₂₂ 

By comparing the corresponding elements of the two matrices,

x = 2 

Thus, 

The value of x = 2.