Question

A square matrix A is said to be orthogonal if \[\text{AA = AA =}{{\text{I}}_{\text{n}}}\] . If A is real skew-
symmetric matrix is such that \[{{\text{A}}^{2}}+I=0\], then
A. A is an orthogonal matrix.
B. A is an orthogonal matrix of odd order.
C. A is an orthogonal matrix of even order.
D. None of these.

Answer 1

Hint:A skew symmetric matrix is a square matrix with its transpose equal to its negative matrix value.

According to the formula, the transpose of A should be equal to negative A and square of matrix A should be equal to the negative of identity matrix and to prove that the matrix is orthogonal we use the matrix formula of:

\[A{{A}^{T}}=I\]

Complete step by step solution:
First as given in the question, the matrix is given as a square matrix with the nature being orthogonal matrix as \[\text{AA = AA =}{{\text{I}}_{\text{n}}}\]

Now a skew symmetric matrix is a matrix in which the transpose of a matrix is negative of the matrix given. With the matrix square being negative that of identity matrix :
\[{{A}^{2}}=-I\]

Now replacing the one of the matrix from the above value to a negative matrix value of \[-A\] with that we have \[A\left( -A \right)=I\].

Now as the skew matrix is negative that of its transpose matrix, replacing the negative matrix to transpose matrix in the above formula, we get the matrix formula as:
\[A\left( {{A}^{T}} \right)=-I\]

The above formula proves that the matrix is orthogonal in nature.
Now to check with the options given if the matrix is even or odd, the only way the above formula remains same when:
\[A\left( {{A}^{T}} \right)={{\left( -1 \right)}^{n}}-I\] with \[n\] being even number as odd number will disprove the orthogonal matrix in nature. Therefore, the matrix \[{{\text{A}}^{2}}+I=0\] is orthogonal and even in number.

Note:A symmetric order matrix is a matrix when transpose of A matrix is equal to the matrix of A and the matrix when become odd is when \[A\left( {{A}^{T}} \right)={{\left( -1 \right)}^{n}}I,n\in \] odd number.