# Consider the matrix A = [[a, b],[c, d]] where det A = ad - bc does not equal 0. Show that the inverse of A is (1/(ad-bc))([[d,-b],[-c, a]]) by reducing [A|I]. You may assume that a does not...

Consider the matrix

A = [[a, b],[c, d]]

where det A = ad - bc does not equal 0. Show that the inverse of A is

(1/(ad-bc))([[d,-b],[-c, a]])

by reducing [A|I]. You may assume that a does not equal 0.

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`([a,b,|,1,0],[c,d,|,0,1])` Multiply row 1 by `1/a`

`==>([1,b/a,|,1/a,0],[c,d,|,0,1])` `R_2->-cR_1+R_2`

`==>([1,b/a,|,1/a,0],[0,(ad-bc)/a,|,-c/a,1])` Multiply row 2 by `a/(ad-bc)`

`==>([1,b/a,|,1/a,0],[0,1,|,(-c)/(ad-bc),a/(ad-bc)])` `R_1->-b/aR_2+R_1`

`==>([1,0,|,d/(ad-bc),-b/(ad-bc)],[0,1,|,(-c)/(ad-bc),a/(ad-bc)])`

**`((-b)/a)((-c)/(ad-bc))+1/a=(bc)/(ad-bc)+1/a=(bc+ad-bc)/(a(ad-bc))=d/(ad-bc)`

Thus if `A=([a,b],[c,d])` then `A^(-1)=1/(ad-bc)([d,-b],[-c,a])`

consider matrix

`A=[[a,b],[c,d]]`

Define

`[A|I]=[[a,b,1,0],[c,d,0,1]]`

`R_1->(1/a)R_1`

`=[[1,b/a,1/a,0],[c,d,0,1]]`

`R_2->R_2-cR_1`

`=[[1,b/a,1/a,0],[0,d-bc/a,-c/a,1]]`

`R_2->(a/(ad-bc))R_2`

`=[[1,b/a,1/a,0],[0,1,-c/(ad-bc),a/(ad-bc)]]`

`R_1->R_1-(b/a)R_2`

`=[[1,0,d/(ad-bc),(-b)/(ad-bc)],[0,1,(-c)/(ad-bc),a/(ad-bc)]]`

`Thus`

`A^(-1)=1/(ad-bc)[[d,-b],[-c,a]]`

`(a!=0, ad-bc!=0``are given)`