Mathematics INVERSE OF A MATRIX
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INVERSE OF A MATRIX

Inverse of a matrix (Reciprocal Matrix) :

Definition: A square matrix `A` said to be invertible (non singular) if there exists a matrix `B` such that,

`AB = I = BA`

`B` is called the inverse (reciprocal) of `A` and is denoted by `A^- 1`. Thus

`A^- 1 = B <=> A B = I = B A`.

Note that for an involutary matrix `A^2 = I => A =A^-1`


`text(Method of finding the inverse of a matrix by Elementary transformation :)`

Let `A` be a non sigular matrix of order `n`. Then A can be reduced to the identity matrix In by a finite sequence of elementary transfonnation only. As we have discussed every elementary row transfonnation of a matrix is equivalent to pre-multiplication by the corresponding elementaty matrix. Therefore there exist elementay matrices `E_1, E_2, ..... , E_4` such that
`(E_k E_(k-1).................E_2E_1)A=I_n`

`=>(E_kE_(k-1)............E_2E_1)A A^(-1)=I_nA^(-1)` (post multiplying by `A^(-1))`

`=>(E_kE_(k-1)...............E_2E_1)I_n=A^(-1)` ( `:. I_nA^(-1)=A^(-1)` and `A A^(-1)=I_n`)

`=> A^(-1)=(E_k E_(k-1)................E_2E_1)I_n`


`text(Algorithm for finding the inverse of a non singular matrix by elementary row transformations :)`

Let `A` be non-singular matrix of order `n`

Step-I : Write `A = I_nA`

Step-II : Perform a sequence of elementaty row operations successively on `A` on the `LHS` and the pre factor `I_n` on the `RHS` till we obtain the result `I_n= BA`

Step-III : Write `A^(- 1) =B`

The following steps will be helpful to find the inverse of a square matrix of order `3` by using elementary row transformations.

Step-I : Introduce unity at the intersection of first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to first row.

Step-II : After introducting unity at `(1, 1)` place introduce zeros at all other places in first column.

Step-III : Introduce unity at the intersection `2^(nd)` row and `2^(nd)` column with the help of `2^(nd)` and `3^(rd)` row.

Step-IV : Introduce zeros at all other places in the second column except at the intersection of `2^(nd)` and `2^(nd)` column

Step-V : Introduce unity at the intersectionofYd row and third column

Step-VI : Finally introduce zeros at all other places in the third colunm except at the intersection of third row and third column.

Properties of Inverse

`text(Property-1)`
Every invertible matrix possesses a unique inverse.

Proof: Let `A` bean invertible matrix of order `n`. Let `B` and `C` be two inverse of `A`.

Then `AB =BA = I_n` ....................(i)

and `AC =CA =I_o` .......................(ii)

Now, `AB =I_n`

`=>C(AB) = C I_n \ \ \ \ \ \ [text( pre-mulriplying by C)]`

`=> (CA) B = C I_n,\ \ \ \ \ \ [text( by associativity)]`

`=> I_n B = C I_n \ \ \ \ \ \ [ ∵ CA = I_B text(from(ii))]`

`=> B = C\ \ \ \ \ \ \ \ [∵ I_n B = B, C I_n = C]`

Hence an invertible matrix possesses a unique inverse.

`text(Property-2)` : If `A` is an invertible square matrix, then `A^T` is also invertible and `(A^T)^-1 = (A^- 1)^T`

Proof: Since `A` is invertible matrix. Therefore,

`|A| !=0`

`=> |A^T| != 0 \ \ \ \ \ \ [ ∵ |A^T| = |A|]`

`=> A^T` is also invertible.

Now, `A A^- 1 = I_n = A^-1A`

`=>(A A^- 1)^T : (I_n)^T = (A^- 1A)^T`

`=> (A^- 1)^T (A^T) = I_n= A^T (A- I)^T`

`=> (A^ T)^-1 = (A^- 1)^T`

`text(Property-3)`. If `A` is a non-singular matrix, then prove that `| A^- 1| = |A |^-1` i.e. `|A^-1| = 1/(|A|)`

Proof: Since `|A|!= 0`, therefore `A^-1` exists such that `A A^-1 = I = A^-1 A`

`=> |A A^-1 | 1=| 1 |`

`=> |A || A^-1 |= 1\ \ \ \ \ \ [∵ |AB| = |A| |B | and |I| =1]`

`=> |A^-1| = 1/(|A|) \ \ \ \ \ [∵ |A| != 0]`

`text(Property - 4 :)`

(Reversal law) If A and Bare invertible matrices of order `n xx n,` then AB is invertible and `(AB)^(-1) = B^(- 1) A^(- 1)` .

`text(Proof)` It is given that A and Bare invertible matrices. `|A| ne 0` and `|B | ne 0 => |A| |B| ne 0`

`=> |AB ne 0|`
Hence, AB is an invertible matrix Now `(AB) (B^(-1)A^(-1)) = A( BB^(-1))A^(-1)`

`= (AI_n)A^(-1)`

`=AA^(-1)`

`=I_n`
Also, `(B^(-1)A^(-1))(AB) = B^(-1) (A^(-1)A) B`

`= B^(-1)(I_nB)`

`= B^(-1)B`

`= I_n`

Thus , `(AB) (B^(-1)A^(-1)) = I_n = (B^(-1)A^(-1)) (AB)`
Hence , `(AB)^(-1) = B^(-1) A^(-1)`

Note If `A,B,C, ... , Y , Z` are invertible matrices, then

`(ABC ... YZ)^(-1) = Z = ^(-1) Y^(-1) ... C^(-1) B^(-1) A^(-1)`

`text(Property 5 :)`

Let A be an invertible matrix of order n, then A' is also

invertible and `(A' )^(-1) = (A^(-1))'`.

`text(Proof:)` A is invertible matrix

`therefore |A| ne 0 => |A'| ne 0` `[∵ |A| = |A'|]`

Hence, `A^(-1)` is also invertible.

Now, `AA^(-1) =I_n= A(-1)A`

`=> (AA^(-1))' = I_n = (A^(-1)A)'` [by reversal law for transpose]

`=>(A')^(-1) = (A^(-1))'` [by definition of inverse]

`text(Property 6: )`

Let A be an invertible matrix of order n and `k in N,` then

`(A^k)^(-1) = (A^(-1))^k = A^(-k)`

`text(Proof)` We have,


`(A^k)^(-1) = undersettext(repeat k times)(underbrace{(A xx A xx A xx .... xx A)^(-1)})`

`= undersettext(repeat k times)(underbrace{A^(-1) xx A^(-1) xx A^(-1) xx .... xx A^(-1)})` [byreversallaw for inverse]

`=(A^(-1))^k = A^(-k)`

`text(Property 7: )`

Let A be an invertible matrix of order n, then `|A^(-1) | =1/|A|` .

`text(Proof)` We have, `A^(-1)A = I_n`

`therefore` Inverse of `A^(-1) = A`
`=> (A^(-1))^(-1) =A`

`text(Note)` `I_n^(-1) = I_n` as `I_n^(-1)I_n = I_n`

`text(Property 8 :)`

Let A be an invertible matrix of order n, then `|A^(-1) | = 1/|A|`

`text(Proof)` ∵ A is invertible, then `|A| ne 0`

`=> |AA^(-1)| = I_n = A^(-1)A`

`=> |AA^(-1)| = |I_n|`

`=> |A| |A^(-1)| = 1` `[ ∵ |AB| = |A| |B| text(and) |I_n| = 1]`

`=> |A^(-1)| = 1/|A|` `[∵ |A| ne 0]`


`text(Property 9 :)`

Inverse of a non-singular diagonal matrix is a diagonal matrix

i.e., If `A = [(a,0,0),(0,b,0),(0,0,c)] text(and) |A| ne 0 ` then

`A^(-1) = [(1/a,0,0),(0,1/b,0),(0,0,1/c)] `

`text(Note)` The inverse of a non-singular square matrix A of order 2 is obtained by
interchanging the diagonal elements and changing signs of off-diagonal elements and
dividing by `|A|.`

For example,

If ` A = [(a,b),(c,d)]` and `|A| = (ad - bc) ne 0` then
`A^(-1) = 1/(ad-bc) [(d,-b),(-c,a)]`


Remark: (i) Note that `A^-1` exists if `A` is non singular.

(ii) If `A =[(a,b),(c,d)]` and `|A| =1` then `A^-1 = [(d,-b),(-c,a)]`

Note:

(i) The necessary and sufficient condition fora square matrix `A` to be invertible is that `|A| != 0`.

(ii) Inverse of a non singular diagonal matrix dia `(k_1, k_2, k_3 ...... k_n)` is the diagonal matrix diag `(k_1^(-1), k_2^(-1), k_3^(-1) ...k_n^(-1) )`
 
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