# Exam question

1. Jun 22, 2010

### johncena

Q:Prove that a square matrix A is invertible iff A is non singular.
My Ans: Since the inverse of a square matrix is given by,
A^ = (1/|A|)adj.A (Where A^ is A inverse)
If |A|=0, A^ is not defined.
i.e, A^ exist only if A is non singular. In other words, a square matrix A is invertible iff A is non singular.
Conversly, Let |A|=0, i.e., let A be singular.
Hence we conclude that A^ exist only if A is non singular.

I got only 1 out of 5 marks for this answer. What is missing in my answer?

2. Jun 22, 2010

### ohubrismine

Let A be a square invertible matrix.
Then there exist a finite number of elementary operations on A that will transform A to I. That is, E1E2...EnA=I. Hence, there exists an inverse of A, namely E1E2...En. So A is nonsingular.
Thus, if A is invertible, then A is nonsingular.

Conversely, let A be nonsingular.
Then there exists an inverse of A.
Hence there exist a finite number of elementary operations on A such that A is transformed to I. That is A^-1 = E1E2...En and (A^-1)A=I. So A is invertible.
Thus, if A is nonsingular, then A is invertible.

Therefore, A is invertible iff A is nonsingular.

3. Jun 22, 2010