● To find inverse of a matrix A, i.e., `A^(–1)` we shall first define adjoint of a matrix.

`\color{green} ✍️` The adjoint of a square matrix `A = [a_(ij ) ]_(n × n)` is defined as the transpose of the matrix `[A_(ij) ]_(n × n)`,

` => color {red} {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 `color{orange}{"adj A =Transpose of" [ (A_11 , A_12, A_13), ( A_21, A_22, A_23), ( A_31, A_32, A_33) ]}`

`color{orange}{ = [ (A_11 , A_21, A_31), ( A_12, A_22, A_32), ( A_13, A_23, A_33) ] }`

For a square matrix of order 2, given by

`A = [ (a_11, a_12 ), ( a_21, a_22) ]`

`Adj A = [(a_22,-a_12),(-a_21,a_11)]`


`color{orange}{"The adj A can also be obtained by interchanging" a_11 "and" a_22 "and by changing signs of" a_12 "and" a_21}`

`"Singular Matrix :"`

`color{red}{"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.


`"NonSingular Matrix : "`

`color{red}{"A square matrix A is said to be non-singular if "| A | ≠ 0}`

Let `A = [ (1,2), (3,4) ]`. Then ` |A | = | (1,2), (3,4) | = 4 -6 = -2 ≠ 0.

Hence A is a nonsingular matrix

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

`color{red}{A(adj A) = (adj A) A = | A | I}`

where `I` is the identity matrix of order n

`"Proof :"`

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) ]`

`=> "As we know 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) ] = |A| I`


Similarly, we can show `color{orange}{(adj A) A = |A | I}`

Hence `color{orange}{A (adj A) = (adj A) A = | A | I`

Therem 1 : `color{red}{"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 :`color{red}{" The determinant of the product of matrices is equal to product of their respective determinants,"}`
`color{red}{" that is," | AB | = | A | | B | , "where A and B are square matrices of the same order"}`


Theorem 3 : `color{red}{"If A is a square matrix of order n, then" |adj(A)| = |A|^(n – 1)}`.


`"Proof :"` We know that (adj A) `A = |A| I = [ ( |A| , 0, 0 ), ( 0, |A| , 0 ), ( 0 ,0 , |A| ) ]`

Writing determinants of matrices on both sides, we have

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

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

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

i.e. `color{orange}{|(adj A)| = |A|^2}`

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

`color{red}{"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)

`=>` `A (1/(|A| ) adj A) =(1/(|A|) adj A ) A= I`

`=>` `AB = BA =I`, where `B= 1/(|A|) adj A`

Thus A is invertible and `color{green}{A^(–1) = 1/(|A|) adj A}`

So `color{blue}{A^(-1) "exist iff" |A| ne 0}`