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Consistency of Linear System of Equations

Let S be th...

Question

Let $S$ be the set of all column matrices $⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦$ such that $b_{1}, b_{2}, b_{3} ϵ R$ and the system of equation (in real variables)
$- x + 2 y + 5 z = b_{1}$
$2 x - 4 y + 3 z = b_{2}$
$x - 2 y + 2 z = b_{3}$
has at least one solution. Then, which of the following system(s) (in real variables) has/have at least one solution of each $⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦ ϵ S$ ?

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

and

x + 2 y + 6 z = b_{3}

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x + y + 3 z = b_{1}, 5 x + 2 y + 6 z = b_{2}

and

- 2 x - y - 3 z = b_{3}

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- x + 2 y - 5 z = b_{1}, 2 x - 4 y + 10 z = b_{2}

and

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

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

and

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

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Solution

The correct options are
A $x + 2 y + 3 z = b_{1}, 4 y + 5 z = b_{2}$ and $x + 2 y + 6 z = b_{3}$
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 $x + 4 y - 5 z = b_{3}$
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 $α P_{1} + λ P_{2} = P_{3}$
$\Rightarrow P_{1} + 7 P_{2} = 13 P_{3}$
$\Rightarrow b_{1} + 7 b_{2} = 13 b_{3}$
(A) $D \neq 0 \Rightarrow$ unique solution for any $b_{1}, b_{2}, b_{3}$
(B) $D = 0$ but $P_{1} + 7 P_{2} \neq 13 P_{3}$
(C) $D = 0$ Also $b_{2} = - 2 b_{1}, b_{3} = - b_{1}$
Satisfied $b_{1} + 7 b_{2} = 13 b_{3}$ (Actually all three planes are co-incident)
(D) $D \neq 0$ .

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Similar questions

S

⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦

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

\begin{matrix} - x + 2 y + 5 z = b_{1} 2 x - 4 y + 3 z = b_{2} x - 2 y + 2 z = b_{3} \end{matrix}

⎡ ⎢ ⎣ \begin{matrix} b_{1} b_{2} b_{3} \end{matrix} ⎤ ⎥ ⎦ \in S

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

x + 2 y + 3 z = 4, 2 x + 3 y + 4 z = 5, 3 x + 4 y + 5 z = 6

x + 2 y + 3 z = 4

2 x + 3 y + 4 z = 5

3 x + 4 y + 5 z = 6

The system of equations

X+2y+3z=4

2x+3y+4z=5

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