We are given that,
A = non-singular matrix
B = non-singular matrix
Order of A = Order of B
We need to find whether AB is singular or non-singular.
Let us recall the definition of non-singular matrix.
Non-singular matrix, also called regular matrix, is a square matrix that is not singular, i.e., one that has a matrix inverse.
We can say that, a square matrix A is non-singular matrix iff its determinant is non-zero, i.e., |A| ≠ 0.
While a singular matrix is a square matrix that doesn’t have a matrix inverse. Also, the determinant is zero, i.e., |A| = 0.
So,
By definition, |A| ≠ 0 and |B| ≠ 0 since A and B are non-singular matrices.
Let,
Order of A = Order of B = n × n
⇒ Matrices A and B can be multiplied
⇒ A × B = AB
If we have matrices A and B of same order then we can say that,
|AB| = 0 iff either |A| or |B| = 0.
And it is clear that, |A|, |B| ≠ 0.
⇒ |AB| ≠ 0
And if |AB| ≠ 0, then by definition AB is s non-singular matrix.
Thus, AB is a singular matrix.
