cla d lb la d lb la dlb m1adbc adbc 1Im 11madbc Thus cla d

c-la d -lb la d -lb la d-lb -m(1(ad-bc)) (ad-bc) (1-Im) = (1-1m)(ad-bc) Thus, c-la d -lb

Solution

Between line 1 and 2 as well as line 2 and 3 , direct steps are used. I will explain clearly.

We know that determinant of   a b

                              c d      = ad-bc

Then,

after 1 st line,

= a (d-lb) - b (c-la) - m [ c(d-lb) - d (c-la) ]

= ad - alb - bc + alb - m [ cd-clb - dc + dla ]

= [ad-bc] - m[ dla-clb]

= [ad-bc] - lm[ ad-bc]

= (1-lm) (ad-bc)