Let S be the set of all column matrices `[(b_(1)),(b_(2)),(b_(3))]` such that `b_(1), b_(2), b_(2) in R` and the system of equations (in real variables)
`-x+2y+5z=b_(1)`
`2x-4y+3z=b_(2)`
`x-2y+2z=b_(3)`
has at least one solution. The, which of the following system (s) (in real variables) has (have) at least one solution for each `[(b_(1)),(b_(2)),(b_(3))] in S` ?

Options

A. `x+2y +3z =b_(1),4y +5z =b_(2) " and " x+2y+6z =b_(3)`
B. `x+y+3z =b_(1), 5x+2y+6z =b_(2) "and" -2x-y-3z =b_(3)`
C. `-x+2y-5z =b_(1), 2x-4y +10z = b_(2) " and x-2y + 5z =b_(3)`
D. `x+2y+5z =b_(1), 2x +3z =b_(2) " and " x + 4y -5z = b_(3)`

Answer

A, D

Explanation

We have,
`-x+2y +5z = b_(1)`
`2x -4y+3z =b_(2)`
`x-2y + 2z = b_(3)`
has at least one solution.
`therefore D = [{:(-1, " "2, 5),(2, -4, 3), (1,-2,2):}]`
and `D_(1) = D_(2) = D_(3)=0`
`rArr D_(1) = [{:(b_(1), 2, 5),(b_(2), -4, 3), (b_(3),-2,2):}]`
`=-2b_(1)-14b_(2) + 26b_(3) =0`
`rArr b_(1) + 7b_(2) = 13b_(3)" "...(i)`
`(a)D = [{:(1, 2, 3),(0, 4, 5), (1,2,6):}] = 1(24-10) + 1(10-12)`
`=14-2 = 12 ne 0`
Here, `D ne 0 rArr` unique solution for any `b_(1), b_(2) b_(3).`
`(b)D = [{:(1, 1, 3),(5, 2, 6), (-2,-1,-3):}]`
`=1(-6+6)-1(-15+12)+3(-5+4)=0`
For atleast one solution
`D_(1) = D_(2)= D_(3) =0`
`"Now,"D_(1) =[{:(b_(1), 1, 3),(b_(2), 2, 6), (b_(3),-1,-3):}]`
`= b_(1)(-6+6)-b_(2)(-3+3) + b_(3)(6-6)`
`= 0`
`D_(2) =[{:(1, b_(1), 3),(5,b_(2), 6), (-2, b_(3),-3):}]`
` =-b_(1)(-15+12) + b_(2)(-3+6)-b_(3)(6-15)`
`=3b_(1) + 3b_(2) +9b_(3) =0rArr b_(1) + b_(2) + b_(3) =0`
not satisfies the Eqs. (i)
It has no solution.
`(c)D =[{:(-1, " "2, -5),(2,-4, 10), (1, -2,5):}]`
`=-1(-20+20)-2(10-10)-5(-4+4)`
`=0`
Here, `b_(2) =-2b_(1) " and" b_(3) =-b_(1) " satisfies the Eqs.(i) Planes are parallel."`
`(d)D =[{:(1,2, 5),(2,0, 3), (1, 4,-5):}] = 1(0-12)-2(-10-3) + 5(8-0)`
=54
`D ne 0`
It has unique solution for any `b_(1), b_(2),b_(3).`