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.

Options

A. 1.`p=2, q ne 3`.
B. 2.`p ne 2, q ne 3`
C. 3.`p ne 2, q = 3`
D. 4.`p =2, q=3`

Answer

b

Explanation

The given system of equations 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 ne0`, i.e., `p ne 2 and q ne 3`, the system has a unique solution.
If `p=2 or q=3, Delta =0`, then if `Delta_(x)=Delta_(y)= Delta_(z)=0`, the system has infinite solutions and if any one of `Delta_(x), Delta_(y) and Delta_(z) ne 0`, the system has no solution.
Now `Delta_(x)= |{:(8,,p,,6),(5,,2,,q),(4,,1,,3):}|`
`" "= 30 -8q-15p+ 4pq= (4q-15)*(p-2)`
`Delta_(y)= |{:(2,,8,,6),(1,,5,,q),(1,,4,,3):}|`
`" "=-8q+ 8q=0`
`Delta _(z) = |{:(2,,p,,8),(1,,2,,5),(1,,1,,4):}|`
Thus, if `p=2, Delta_(x)=Delta_(y) = Delta_(z)= 0 ` for all `q in R`, the system has infinite solutions.
If `p ne 2, q = 3 and Delta_(z) ne 0`, then the system has no solutions.
Hence the system has (i) no solution if `p ne 2 and q=3`, (ii) a unique solution if `p ne 2 and q ne 3` and (iii) infinite solutions of `p=2 and q in R`.