Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_{1} \\ b_{2} \\ b_{3}\end{array}\right]$ such that $b_{1}, b_{2}, b_{3} \in \mathbb{R}$ and the system of equations (in real variables) \[ \begin{aligned} -x+2 y+5 z & =b_{1} \\ 2 x-4 y+3 z & =b_{2} \\ x-2 y+2 z & =b_{3} \end{aligned} \] has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $\left[\begin{array}{l}b_{1} \\ b_{2} \\ b_{3}\end{array}\right] \in S$?

(A) $x+2 y+3 z=b_{1}, 4 y+5 z=b_{2}$ and $x+2 y+6 z=b_{3}$; (D) $x+2 y+5 z=b_{1}, 2 x+3 z=b_{2}$ and $x+4 y-5 z=b_{3}$

(B) $x+y+3 z=b_{1}, 5 x+2 y+6 z=b_{2}$ and $-2 x-y-3 z=b_{3}$

(C) $-x+2 y-5 z=b_{1}, 2 x-4 y+10 z=b_{2}$ and $x-2 y+5 z=b_{3}$

Matrices — JEE Advanced · 2018 · 20 May 2018 · Paper 2 Question with Solution

Let

S

be the set of all column matrices

b_{1} b_{2} b_{3}

such that

b_{1}, b_{2}, b_{3} \in R

and the system of equations (in real variables)

- x + 2 y + 5 z 2 x - 4 y + 3 z x - 2 y + 2 z = b_{1} = b_{2} = b_{3}

has at least one solution. Then, 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

(A) $x + 2 y + 3 z = b_{1}, 4 y + 5 z = b_{2}$ and $x + 2 y + 6 z = b_{3}$ ✓ Correct
(B) $x + y + 3 z = b_{1}, 5 x + 2 y + 6 z = b_{2}$ and $- 2 x - y - 3 z = b_{3}$
(C) $- x + 2 y - 5 z = b_{1}, 2 x - 4 y + 10 z = b_{2}$ and $x - 2 y + 5 z = b_{3}$
(D) $x + 2 y + 5 z = b_{1}, 2 x + 3 z = b_{2}$ and $x + 4 y - 5 z = b_{3}$ ✓ Correct

Answer: (A), (D)

The determinant of the coefficient matrix is:

Δ = - 1 21 2 - 4 - 2 532 = 0

So the given equations represent a family of planes. For consistency, we find

13 b_{3} = b_{1} + 7 b_{2}

.

(A):

101242356 \neq = 0

, so the system has a unique solution for any

b_{1}, b_{2}, b_{3}

. ✓

(B): The determinant is 0, and checking consistency shows it requires

b_{1} + b_{2} + 3 b_{3} = 0

, which is not always true for elements in

S

.

(C): The three equations represent parallel planes, which only have a solution in specific cases.

(D):

121204 53 - 5 \neq = 0

, so the system has a unique solution for any

b_{1}, b_{2}, b_{3}

. ✓

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options