# Adjoint and Inverse of Matrix for Class 12/center>

• 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 $a_{ij}$ .
Let $A= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$
$Adj A = transpose \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$

## Theorem of Adjoint of Matrix

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

## Singular Matrix

A square matrix A is said to be singular if |A| = 0
For a 2x2 matrix $A$ given by
$A = \begin{vmatrix} 1 & 2 \\ 4 & 8 \end{vmatrix}=0$

## Non Singular Matrix

A square matrix A is said to be non-singular if $|A| \ne 0$
$A = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}=-2$

## More Theorems

Theorem 1:
If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order.

Theorem 2:
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

Thoerem 3:
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
|AB|=|I|
From theorem 2
|A||B|=1
Hence $|A| \ne 0$
So A is a non-singular

## How to Find Invertible Matrix with the Adjoint

We know from previous theorem
If A be any given square matrix of order n, then
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 NCERT Solutions & Assignments

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