Adjoint and Inverse of Matrix for Class 12/center>
Adjoint of Matrix
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
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
A(adj A) = (adj A) A = |A| I 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
Hence A (adj A) = (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
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 $A^{-1} = \frac {1}{|A|} adj A$
