A square matrix, which has its inverse is called an invertible or non-singular matrix. If matrix A has its inverse, i.e AB = I = BA, then A is called invertible matrix and B is the inverse of matrix A. If a square matrix is non invertible i.e. inverse of that matrix does not exists then such type of matrix is called as singular matrix and we know that the determinant of singular matrices is equal to zero(0).

If Amxm order matrix, then the order of inverse matrix of A i.e. order of A-1 is mxm and if A is invertible then it has one and only one inverse matrix and that inverse matrix is again invertible.

If Amxm order square matrix, then the below statements are corresponds to each other
( either all true or false).
  • A-1 exists, so A is invertible.
  • The value of the determinant of A is zero, i.e. IAI $\neq$ o.
  • None of the eigen value of A is zero.
  • The transpose matrix of A is also invertible.
  • System of linear equations, having A as the coefficient matrix(Ax = B), has atleast one solution.
  • Rank of A is exists.
  • After applying elementary row operations, A converted to an identity matrix.
  • Homogeneous system of equations Ax = 0, has trivial solution.
  • There exists a matrix B of order mxm, such that AB = I = BA.

Example 1: A = $\begin{bmatrix}
7 &9 \\
4 &3
\end{bmatrix}$, show that A is invertible matrix and calculate its inverse. Further prove result also.

Solution: If A is invertible then IAI $\neq$ 0.

So IAI = (21 - 39 = -15 $\neq$ 0

Hence A is invertible.
Now A-1 = $\frac{adj A}{IAI}$

For adj A, first we have to calculate cofactor matrix say C of A, then

C = $\begin{bmatrix}
3 &-4 \\
-9&7
\end{bmatrix}$

So adj A = CT = $\begin{bmatrix}
3 &-9 \\
-4 &7
\end{bmatrix}$

Hence A-1 = $\frac{\begin{bmatrix}
3 &-9 \\
-4 &7
\end{bmatrix}}{-15}$

= $\frac{1}{-15}$ . $\begin{bmatrix}
3 &-9 \\
-4 &7
\end{bmatrix}$

Again ( A A-1) = $\begin{bmatrix}
7 &9 \\
4 &3
\end{bmatrix}$ . $\frac{1}{-15}$ . $\begin{bmatrix}
3 &-9 \\
-4 &7
\end{bmatrix}$

= $\frac{1}{-15}$ . $\begin{bmatrix}
7 &9 \\
4 &3
\end{bmatrix}$. $\begin{bmatrix}
3 &-9 \\
-4 &7
\end{bmatrix}$

= $\frac{1}{-15}$ . $\begin{bmatrix}
-15&0 \\
0 &-15
\end{bmatrix}$

= $\begin{bmatrix}
1&0 \\
0 &1
\end{bmatrix}$

So A.A-1 = I.

It means, if we multiply of a matrix and is't inverse then we get identity matrix as the result.

In matrices, if a square matrix is singular( determinant of matrix is zero) then the inverse of such type of matrix is not exists. Then that matrix is called as non-invertible matrix. If we have non-square matrices i.e rectangle matrices, then we don't find its inverse because the inverse is exists only when the matrix should be square.

Example 1
: If A = $\begin{bmatrix}
2 &7 \\
-2&-7
\end{bmatrix}$ be a square matrix, find the inverse of it.

Solution: For inverse of the matrix, the determinant of matrix is non zero. So first we have
to calculate determinant of A,
IAI = (-14 + 14 ) = 0
So inverse of A does not exists.

Example2
: If A = $\begin{bmatrix}
1 &3 &-5 \\
0 &-6 &7
\end{bmatrix}$, then find its inverse.

Solution: Given matrix A has the order 2x3 and for this rectangular matrix we can't calculate
its determinant( only square matrices has its determinant values). So inverse of
matrix A not exists.

It may be noted that the inverse is exists only for square matrices, non square matrices do not have its inverse.