1. All diagonal elements are zero.
2. The sum of all the diagonal elements of the matrix is zero.
3. \(A\) is orthogonal matrix.
Which of the statements given above are correct?
A skew-symmetric matrix is a square matrix \(A\) where its transpose (\(A^T\)) is equal to its negative (\(-A\)). This relationship can be written as:
\(A^T = -A\)
This definition implies that for any element \(a_{ij}\) (the element in the \(i\)-th row and \(j\)-th column), the element \(a_{ji}\) must satisfy \(a_{ji} = -a_{ij}\). We need to check three statements regarding a \(3 \times 3\) skew-symmetric matrix.
Statement 1 says that all diagonal elements of the matrix are zero.
For any diagonal element, the row index is the same as the column index, meaning \(i=j\). So, we are looking at elements \(a_{ii}\).
Using the property \(a_{ji} = -a_{ij}\) for \(i=j\), we get:
\(a_{ii} = -a_{ii}\)
If we add \(a_{ii}\) to both sides of the equation, we get:
\(2a_{ii} = 0\)
Dividing by 2 yields:
\(a_{ii} = 0\)
This confirms that all diagonal elements (\(a_{11}, a_{22}, a_{33}\)) must be zero. So, Statement 1 is correct.
Statement 2 claims that the sum of all the diagonal elements is zero.
The sum of the diagonal elements of a matrix is called its trace. For a \(3 \times 3\) matrix \(A\), the trace is \(tr(A) = a_{11} + a_{22} + a_{33}\).
From Statement 1, we established that \(a_{11} = 0\), \(a_{22} = 0\), and \(a_{33} = 0\). Therefore, the sum is:
\(tr(A) = 0 + 0 + 0 = 0\)
This means the sum of the diagonal elements is zero. So, Statement 2 is correct.
Statement 3 proposes that the skew-symmetric matrix \(A\) is an orthogonal matrix.
A matrix is orthogonal if its transpose equals its inverse, satisfying \(A^T A = I\), where \(I\) is the identity matrix.
For a skew-symmetric matrix, we know \(A^T = -A\). Substituting this into the orthogonality condition:
\((-A)A = I\)
\(-A^2 = I\)
\(A^2 = -I\)
Let's consider a general \(3 \times 3\) skew-symmetric matrix:
| \(A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}\) |
Calculating \(A^2\):
\(A^2 = A \times A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}\)
\(A^2 = \begin{pmatrix} -a^2-b^2 & -bc & ac \\ -bc & -a^2-c^2 & -ab \\ ac & -ab & -b^2-c^2 \end{pmatrix}\)
For \(A\) to be orthogonal, \(A^2\) must equal \(-I\):
\(\begin{pmatrix} -a^2-b^2 & -bc & ac \\ -bc & -a^2-c^2 & -ab \\ ac & -ab & -b^2-c^2 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\)
This requires the off-diagonal terms to be zero (\(ac=0, -bc=0, ab=0\)) and the diagonal terms to be -1 (\(-a^2-b^2 = -1\), \(-a^2-c^2 = -1\), \(-b^2-c^2 = -1\)).
If \(a=0\), we need \(b^2=1\), \(c^2=1\), and \(b^2+c^2 = 1\). But \(1+1 = 2 \neq 1\). This shows a contradiction. A general skew-symmetric matrix does not satisfy \(A^2 = -I\). For example, the zero matrix is skew-symmetric but \(0^2 \neq -I\). Therefore, Statement 3 is incorrect.
Reviewing the analysis:
Thus, the correct statements are 1 and 2 only.