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Linear System of Equations

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Question

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

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 option is D $x + 2 y + 5 z = b_{1}, 2 x + 3 z = b_{2} and x + 4 y - 5 z = b_{3}$
$\begin{matrix} - x + 2 y + 5 z = b_{1} . . . . . . . . . . (1) 2 x - 4 y + 3 z = b_{2} . . . . . . . . . . . . (2) x - 2 y + 2 z = b_{3} . . . . . . . . . . . . (3) \end{matrix}$

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 2 & 5 2 & - 4 & 3 1 & - 2 & 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$
No two planes are parallel so infinite solutions.
They are family of planes so,
$P_{3} = λ P_{1} + μ P_{2} 1 = - λ + 2 μ . . . . . . . (i) - 2 = 2 λ - 4 μ . . . . . . . (i i) 2 = 5 λ + 3 μ . . . . . . . (i i i)$
Using equation $(i)$ and $(i i i)$ ,
$λ = \frac{1}{13}; μ = \frac{7}{13} \Rightarrow 13 b_{3} = b_{1} + 7 b_{2} . . . . . . (1)$

Now checking $Δ$ for all the option,
Option (a)
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 3 0 & 4 & 5 1 & 2 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 12 \Rightarrow Δ \neq 0$

Option (b)
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 3 5 & 2 & 6 - 2 & - 1 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$
So
$P_{3} = λ P_{1} + μ P_{2} - 2 = λ + 5 μ . . . . . . . (i v) - 1 = λ + 2 μ . . . . . . . (v) - 3 = 3 λ + 6 μ . . . . . . . (v i)$
Using equation $(i v)$ and $(v)$ ,
$λ = - \frac{1}{3}; μ = - \frac{1}{3}$
So $- \frac{b_{1}}{3} - \frac{b_{2}}{3} = b_{3} \Rightarrow - 3 b_{3} = b_{1} + b_{2}$
Which is not equal to equation $(1)$
No solutions.

Option (c)
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} - 1 & 2 & - 5 2 & - 4 & 10 1 & - 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Directly from the equation we can wrie
$b_{1} = - b_{3}$ and $b_{2} = 2 b_{3}$
Putting this in equation $(1)$
$14 b_{3} - b_{3} = 13 b_{3}$
satisfies the equation, so infinite solution.

Option (d)
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 5 2 & 0 & 3 1 & 4 & - 5 \end{matrix} ∣ ∣ ∣ ∣ = 54 \neq 0$

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Similar questions

S

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

b_{1}, b_{2}, b_{3} ϵ R

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

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

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

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

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

x + 2 y + 3 z = 4

2 x + 3 y + 4 z = 5

3 x + 4 y + 5 z = 6

2 x - 4 y + 3 z = 1, x - 2 y + 4 z = 3, 3 x - y + 5 z = 2

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