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Question:

Let S be the set of all column matrices $\begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}$ such that $b_1, b_2, b_3 \in \mathbb{R}$ and the system of equations (in real variables) $-x + 2y + 5z = b_1$, $2x - y + 3z = b_2$, $x - y + 2z = b_3$ has at least one solution. Then, which of the following system(s) (in real variables) has/have at least one solution for each $\begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix} \in S$?

-x + 2y - z = b_1, 2x - y + 10z = b_2 and x - y + 5z = b_3

x + 2y + 5z = b_1, 2x + 3z = b_2 and x + 4y - z = b_3

x + y + 3z = b_1, 5x + 2y + 6z = b_2 and -x - y - z = b_3

x + 2y + 3z = b_1, 4y + 5z = b_2 and x + 2y + 6z = b_3

Solution:

We find D = 0, where D is the determinant formed by the coefficients of x, y, z in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions. Let αP1 + λP2 = P3 ⇒ P1 + 7P2 = 13P3 ⇒ b1 + 7b2 = 13b3 (A) D ≠ 0 ⇒ unique solution for any b1, b2, b3 (B) D = 0 but P1 + 7P2 ≠ 13P3 (C) D = 0 Also b2 = -b1, b3 = -b1 Satisfied b1 + 7b2 = 13b3 (Actually all three planes are co-incident) (D) D ≠ 0.