Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

Defination:  The adjoint of a square matrix A = ${\displaystyle {\left[a_{ij}\right]}_{n\times n}}$ is defined as the transpose of the matrix ${\displaystyle {\left[A_{ij}\right]}_{n\times n}}$, where ${\displaystyle A_{ij}}$ is the cofactor of the element ${\displaystyle a_{ij}}$. Adjoint of the matrix A is denoted by adj A.
Let A = ${\displaystyle \left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}$

Then adj A = Transpose of ${\displaystyle \left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]}$

If A be any given square matrix of order n, then
A(adj A) = (adj A) A = |A| I , 
where I is the identity matrix of order n
Verification 
Let A =${\displaystyle \left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]}$ , then adj A =${\displaystyle \left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]}$
Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have
A (adj A) =  ${\displaystyle \left|\begin{array}{ccc}\left|A\right|& 0& 0\\ 0& \left|A\right|& 0\\ 0& 0& \left|A\right|\end{array}\right|=\left|A\right|\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$
Similarly, we can show  (adj A) A = A I
Hence A (adj A) = (adj A) A = A I

Definition :  A square matrix A is said to be singular if A = 0.
For example, the determinant of matrix A = ${\displaystyle \left[\begin{array}{cc}1& 2\\ 4& 8\end{array}\right]}$ is zero
Hence A is a singular matrix.

Definition :  A square matrix A is said to be non-singular if A ≠ 0
Let A = ${\displaystyle \left[\begin{array}{cc}1& 4\\ 3& 2\end{array}\right]}$. Then |A| = ${\displaystyle \left|(1,4)(3,2)\right|}$ = 4 - 6 = 2 ≠ 0
Hence A is a nonsingular matrix.
We state the following theorems without proof.

Theorem : If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order.
Theorem: The determinant of the product of matrices is equal to product of their respective determinants, that is, |AB| = |A| |B| , where A and B are square matrices of the same order
Remark:  We know that (adj A) A = |A| I = ${\displaystyle \left[\begin{array}{ccc}\left|A\right|& 0& 0\\ 0& \left|A\right|& 0\\ 0& 0& \left|A\right|\end{array}\right]}$ , |A| ≠ 0
Writing determinants of matrices on both sides, we have
${\displaystyle \left|\left(adjA\right)A\right|=\left|\begin{array}{ccc}\left|A\right|& 0& 0\\ 0& \left|A\right|& 0\\ 0& 0& \left|A\right|\end{array}\right|}$

i.e,. ${\displaystyle \left|\left(adjA\right)\right|\left|A\right|={\left|A\right|}^3\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|}$

i.e. ${\displaystyle \left|\left(adjA\right)\right|\left|A\right|=\left|A\right|3\left(1\right)i.e.\left|\left(adjA\right)\right|=\left|A\right|2}$

i.e. ${\displaystyle \left|\left(adjA\right)\right|={\left|A\right|}^2}$

In general, if A is a square matrix of order n, then |adj(A)| = |A|^(n – 1).

Theorem:  A square matrix A is invertible if and only if A is nonsingular matrix.

Proof:  Let A be invertible matrix of order n and I be the identity matrix of order n. Then, there exists a square matrix B of order n such that 
AB = BA = I
Now   AB = I.  So |AB| = I or |A| |B| = 1  (since |I| =1, |AB|=|A||B|)
This gives |A| ≠ 0.
Hence A is nonsingular.
Conversely, let A be nonsingular. Then A ≠ 0
Now A (adj A) = (adj A) A = |A| I                  (Theorem 1)
or ${\displaystyle A\left(\frac{1}{\left|A\right|}adjA\right)=\left(\frac{1}{\left|A\right|}adjA\right)A=I}$
or ${\displaystyle AB=BA=I}$ ,where  ${\displaystyle B=\frac{1}{\left|A\right|}adjA}$
Thus, A is invertible and ${\displaystyle A^{-1}=\frac{1}{\left|A\right|}adjA}$