Invertible Matrices




If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by `A^(– 1)`. In that case A is said to be invertible.
For example, let  
A = `[(2,3),(1,2)]` and B = `[(2,-3),(-1,2)]` be two matrices.

Now AB =`[(2,3),(1,2)][(2,-3),(-1,2)]`

=`[(4-3,-6+6),(2-2,-3+4)]`=`[(1,0),(0,1)]` = I

Also BA = `[(1,0),(0,1)]` = I .Thus B is the inverse of A , in order words B =` A^(–1)` and A is inverse of B, i.e., A =` B^(–1)`.


(Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique. 
Proof:  Let A = `[a_(ij)]` be a square matrix of order m. If possible, let B and C be two inverses of A. We shall show that B = C. 
Since B is the inverse of A 
AB = BA = I                  ... (1) 
Since C is also the inverse of A 
AC = CA = I                  ... (2) 
Thus   B = BI = B (AC) = (BA) C = IC = C  


If A and B are invertible matrices of the same order, then `(AB)^(–1) = B^(–1) A^(-1)`.

Proof: From the definition of inverse of a matrix, we have 
`(AB) (AB)^(–1) = 1`

or `A^(–1) (AB) (AB)^(–1) = A^(–1)I`   
(Pre multiplying both sides by `A^(–1)`) 
or `(A^(–1)A) B (AB)^(–1)` = `A^(–1)`
(Since `A^(–1) I = A^(–1)`)

or `IB (AB)^(–1) = A^(–1)`

or `B (AB)^(–1) = A^(–1)`

or `B^(–1) B (AB)^(–1) = B^(–1) A^(–1)`

or `I (AB)^(–1) = B^(–1) A^(–1)`

Hence `(AB)^(–1) = B^(–1) A^(–1)`

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