For what values of p and q the system of equations `2x+py+6z=8, x+2y+qz=5, x+y+3z=4` has i no solution ii a unique solution iii in finitely many solutions.

Explanation

the given system of equation is
`2x+py+ 6z=8`
`x+2y+qz=5`
`x+y +3z=4`
`Delta = |{:(2,,p,,6),(1,,2,,q),(1,,1,,3):}| =(2-p)(3-q)`
By Cramer's rule if `Delta ne 0` i.e., p `ne 2, qne 3` the system has unique solution
if p=2 or q=3 ,`Delta =0` , then if `Delta_(x)=Delta_(z)=0` the system has infinite solutions and if at least one of `Delta_(x),Delta_(y),Delta_(z) ne0` systems has no solution Now
`|{:(8,p,6),(5,2,q),(4,1,4):}|`
`=30 -3q -15q +4pq+(p-2)(4q-15)`
`Delta _(y)= |{:(2,,8,,6),(1,,5,,q),(1,,4,,3):}`|
`=-8q+8q=0`
`Delta _(z)= |{:(2,,q,,8),(1,,2,,5),(1,,1,,4):}|`
`=p-2`
Thus if p =2 `Detla _(x)=Delta_(y)=Delta_(z)=0` for all `q in R` so the system has infinite solution ,
And if `p ne 2, q=3 Delta _(x), Delta_(z) ne0` the system has no solution Hence the system has
(i) no solution if `p ne 2, q=3`
(ii) a unique solution ` if p ne 2, qne 3`
(iii) infinitely many solutions `p=2 , q in R`