Let S be the of all column matrices b1b2b3 such that b1, b2, b3∈ℝ and the system of equations (in real variables) -x+2y+5z=b12x-4y+3z=b2x-2y+2z=b3 has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution of each b1b2b3ϵ S ?

(A) x+2y+3z=b1, 4y+5z=b2 and x+2y+6z=b3

(B) x+y+3z=b1, 5x+2y+6z=b2 and -2x-y-3z=b3

(C) –x+2y-5z=b1, 2x-4y+10z=b2 and x-2y+5z=b3

JEE Advanced · Mathematics · 19. Determinants

Let $S$ be the 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 of each $[\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix}] ϵ S$ ?

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

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

We find

D = 0

& since no pair of planes are parallel, so there are an infinite number of solutions. So, the infinite solutions shall lie on a common line of intersection of these planes. Hence, we can write any plane as a linear combination of other two planes:

α P_{1} - λ P_{2} = P_{3}

\Rightarrow

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

(by comparing coefficients of

x, y, z

)

\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 as they are parallel planes, they will have at least one solution only if they are all coincident, so that

b_{2} = - 2 b_{1}, b_{3} = - b_{1}

So each of

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

that satisfy

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

, they don't always need to satisfy

b_{2} = - 2 b_{1}, b_{3} = - b_{1}

. Hence option is wrong.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

If the adjoint of a $3 \times 3$ matrix $P$ is $\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$, then the possible value(s) of the determinant of $P$ is (are)

JEE Advanced 2012 Easy

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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Let S be the of all column matrices b1b2b3 such that b1, b2, b3∈ℝ and the system of equations (in real variables)-x+2y+5z=b12x-4y+3z=b2x-2y+2z=b3has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution of each b1b2b3ϵ S ?

Step-by-step Solution