If A

_{mxm} 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 = C^{T} = $\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.