- The adjoint of a square matrix $A = [a_{ij}] _{n \times n}$ is defined as the transpose of the matrix $[A_{ij}]_{n \times n}$, where $A_{ij}$ is the cofactor of the element $a_{ij}$ .
- Adjoint of the matrix A is denoted by adj A

$Adj A = transpose \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} $

A(adj A) = (adj A) A = |A| I

Let \[ A= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

$Adj A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} $

Now

$A(adj A)=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} $

Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have

$=\begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix}= |A|\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =|A| I$

Similarly, we can show (adj A) A = |A| I

Hence A (adj A) = (adj A) A = |A| I

For a 2x2 matrix \( A \) given by

\[ A = \begin{vmatrix} 1 & 2 \\ 4 & 8 \end{vmatrix}=0 \]

A square matrix A is said to be non-singular if $|A| \ne 0$

\[ A = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}=-2 \]

If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order.

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

A square matrix A is invertible if and only if A is nonsingular matrix

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

|AB|=|I|

From theorem 2

|A||B|=1

Hence $|A| \ne 0$

So A is a non-singular

If A be any given square matrix of order n, then

A(adj A) = (adj A) A = |A| I

Now if we assume that A is non sigular, then $|A| \ne 0$

So dividing whole equation by |A|

$A( \frac {1}{|A|} adj A)= ( \frac {1}{|A|} adj A) A = I$

or AB = BA = I where $B= \frac {1}{|A|} adj A$

So $A^{-1} = \frac {1}{|A|} adj A$

**Notes**- What is Determinants
- Properties of Determinants
- Minor and Cofactors of Determinants
- Adjoint and Inverse of Matrix

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