Find the value of `a \ a n d \ b` if the system of equation `a^2x-b y=a^2-b \ a n d \ b x-b^2y=2+4b` (i) posses unique solution (ii) infinite solutions

Explanation

System of equations is
`a^(2)x-by=a^(2)-b" and " bx-b^(2)y =2 +4b`
(i) if system has unique solution then lines must be non- parallel or `Delta ne 0` Hence,
`|{:( a^(2),,-b),(b,,-b^(2)):}| ne0`
`" or " -a^(2)b^(2)+b^(2)ne0`
` " or " b^(2) (1-a^(2)) ne0`
` rArr bne " and "ane=1`
(ii) if system has infinite solutions then lines must be coincident. Hence
`(a^(2))/(b)=(b)/(b^(2)) =(a^(2)-b)/(2+4b)`
`rArr b=0 " or " a = ne 1`
`" if " a=1 " then " 2 + 4b =b-b^(2)`
`" or " b^(2)+3b+2=0`
`rArr b=-2 " or " -1`
`" if " a=-1 , " then " b=-2 " or " -1`
b=0 is not possible
Then ordered pairs (a,b) for which system has infintie solutions are `( 1,-2) (1,-1)(-1,-2)(-1,-1)`