#### definition

**Defination:** The adjoint of a square matrix A = `[a_(ij)]_(n × n)` is defined as the transpose of the matrix `[A_(ij)]_(n × n)`, where `A_(ij)` is the cofactor of the element `a_(ij)`. Adjoint of the matrix A is denoted by adj A.

Let A = `[(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)]`

Then adj A = Transpose of `[(A_11,A_12,A_13),(A_21,A_22,A_23),(A_31,A_32,A_33)] = [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`

#### theorem

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 =`[(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)]` , then adj A =` [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`

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) = `|(|A|,0,0),(0,|A|,0),(0,0,|A|)| = |A| [(1,0,0),(0,1,0),(0,0,1)]`

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

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

#### notes

**Definition :** A square matrix A is said to be singular if A = 0.

For example, the determinant of matrix A = `[(1,2),(4,8)]` is zero

Hence A is a singular matrix.

**Definition :** A square matrix A is said to be non-singular if A ≠ 0

Let A = `[(1,4),(3,2)]`. Then |A| = `|(1,4)(3,2)|` = 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 = `[(|A|,0,0),(0,|A|,0),(0,0,|A|)]` , |A| ≠ 0

Writing determinants of matrices on both sides, we have

`|(adj A)A| = |(|A|,0,0),(0,|A|,0),(0,0,|A|)|`

i.e,. `|(adjA)| |A| = |A|^3 |(1,0,0),(0,1,0),(0,0,1)|`

i.e. `|(adj A)| |A| = |A|3 (1) i.e. |(adj A)| = |A|2`

i.e. `|(adj A)| = |A|^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 `A(1/|A| adj A) = (1/|A| adj A) A = I`

or `AB = BA = I` ,where `B = 1/|A| adj A`

Thus, A is invertible and `A^-1 = 1/ |A|adj A`