Given, AB = BA
`[[1,2],[3,4]] [[a,b],[c,d]]=[[a,b],[c,d]][[1,2],[3,4]]`
`rArr[[a+2c,b+2d],[3a+4c,3b+4d]]=[[a+3b,2a+4b],[c+3d,2c+4d]] `
On comparing, we get
`a + 2c = a +3b`
`rArr b= (2c)/3` …(i)
`b + 2d =2a +2b`
`rArr d = a+(3b)/2 ` ... (ii)
`rArr 3a + 4c= c + 3d`
`d = a +c` ...(iii)
and `3b + 4 d = 3c + 4d`
`rArr b = 2c/3` ...(iv)
`rArr (d-b)/(a+c-b)=(d-b)/(d-b) = 1 ` [frp, Eq. (iii)] Now, `B= [[a,b],[c,d]]= [[d-c,(2c)/3],[c,d]]`
If `c= beta and d= alpha,` then `B= [[alpha-beta ,(2beta)/3],[beta,alpha]]`