# 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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Given matrix is `A=[[a,b],[c,d]]` .

To find the inverse of the matrix A we can write

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

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

Which is the desired result.

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

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

`R_1->(1/a)R_1`

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

`R_2->R_2-cR_1`

[A|I]=`[[1,b/a,1/a,0],[0,d-(cb)/a,-c/a,1]]`

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

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

[A|I]=`[[1,b/a,1/a,0],[0,1,-c/(ad-cb),a/(ad-cb)]]`

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

`[A|I]=[[1,0,1/a+(bc)/(a(ad-cb)),-b/(ad-cb)],[0,1,-c/(ad-cb),a/(ad-cb)]]`

`[A|I]=1/(ad-cb)[[1,0,d,-b],[0,1,-c,a]]`

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