Which of the following statement is false?a)Every skew symmetric matri...
False statement: Every skew symmetric matrix of odd order is non-singular.
Explanation:
To understand why this statement is false, let's first define what a skew symmetric matrix is and what it means for a matrix to be non-singular.
A skew symmetric matrix is a square matrix where the elements below the main diagonal are the negatives of the corresponding elements above the main diagonal. In other words, for any skew symmetric matrix A, A[i][j] = -A[j][i] for all i and j.
A matrix is said to be non-singular (or invertible) if its determinant is non-zero. The determinant of a matrix is a scalar value that can be calculated using various methods, such as cofactor expansion or row reduction.
Counterexample:
Let's consider a counterexample to prove that the statement is false. Suppose we have a 3x3 skew symmetric matrix A:
A = [0 a b]
[-a 0 c]
[-b -c 0]
where a, b, and c are non-zero real numbers.
Now, let's calculate the determinant of matrix A:
det(A) = 0(a(0) - c(-b)) - a(-a(0) - c(-b)) + b(-a(-b) - c(0))
= 0 + a^2 + b^2
Since a, b, and c are non-zero, the determinant of matrix A is non-zero. Therefore, matrix A is non-singular.
From this counterexample, we can see that there exist skew symmetric matrices of odd order that are non-singular, which contradicts the given statement. Hence, the statement is false.
Conclusion:
The false statement is: Every skew symmetric matrix of odd order is non-singular.