Let x∈R and let P=111022003, Q=2xx040xx6 and R=PQP-1. Then which of the following options is/are correct?

(D) For x=0, if R1ab=61ab, then a+b=5

(A) For x=1, there exists a unit vector αi^+βj^+γk^ for which Rαβγ=000

(B) There exists a real number x such that PQ=QP

(C) det⁡R=det⁡2xx040xx5+8, for all x∈R

(D) For x=0, if R1ab=61ab, then a+b=5

JEE Advanced · Mathematics · 19. Determinants

Let $x \in R$ and let $P = [\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}], Q = [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{matrix}]$ and $R = P Q P^{- 1} .$
Then which of the following options is/are correct?

A For $x = 1,$ there exists a unit vector $α \hat{i} + β \hat{j} + γ \hat{k}$ for which $R [\begin{matrix} α \\ β \\ γ \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$
B There exists a real number $x$ such that $P Q = Q P$
C $\det R = \det [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{matrix}] + 8,$ for all $x \in R$
D For $x = 0,$ if $R [\begin{matrix} 1 \\ a \\ b \end{matrix}] = 6 [\begin{matrix} 1 \\ a \\ b \end{matrix}],$ then $a + b = 5$

Verified Solution

Answer & Solution

Correct Answer

(D) For $x = 0,$ if $R [\begin{matrix} 1 \\ a \\ b \end{matrix}] = 6 [\begin{matrix} 1 \\ a \\ b \end{matrix}],$ then $a + b = 5$

Step-by-step Solution

Detailed explanation

\begin{matrix} P = [\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}] & Q = [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{matrix}] \end{matrix}

Option

(C)

Now

R = P Q P^{- 1}

\det (R) = \det (R) \det (Q) \det (P^{- 1})

\begin{matrix} \det (R) = \det (Q) & (\det (P^{- 1}) = \frac{1}{\det (P)}) \end{matrix}

\det (R) = \det [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{matrix}]

\det (R) = 48 - 4 x^{2}

now

\det [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{matrix}] = 40 - 4 x^{2}

\det R = \det [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{matrix}] + 8

Option

(A)

R [\begin{matrix} α \\ β \\ γ \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

must have not trivial solution
So

\det (R) = 0

48 - 4 x^{2} = 0 \Rightarrow x = \pm 2 \sqrt{3}

Option

(D)

R [\begin{matrix} 1 \\ a \\ b \end{matrix}] = [\begin{matrix} 6 \\ 6 a \\ 6 b \end{matrix}]

P Q P^{- 1} [\begin{matrix} 1 \\ a \\ b \end{matrix}] = [\begin{matrix} 6 \\ 6 a \\ 6 b \end{matrix}]

…(i)

P^{- 1} = \frac{1}{6} [\begin{matrix} 6 & - 3 & 0 \\ 0 & 3 & - 2 \\ 0 & 0 & 2 \end{matrix}]

Putting

P, Q, P^{- 1}

in equation (i)

\frac{1}{6} [\begin{matrix} 12 & 6 & 4 \\ 0 & 24 & 8 \\ 0 & 0 & 36 \end{matrix}] [\begin{matrix} 1 \\ a \\ b \end{matrix}] = [\begin{matrix} 6 \\ 6 a \\ 6 b \end{matrix}]

\Rightarrow 12 + 6 a + 4 b = 36

24 a + 8 b = 36 a

a = 2

and

b = 3

a + b = 5

Option

(B)

P Q = Q P

P Q = Q P \Rightarrow P Q P^{- 1} = Q

R = Q

Not possible for any value of

x .

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.

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Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right) ; k=1,2, \ldots, 9$

List - I	List - II
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(B) There exists a $k \in\{1,2, \ldots, 9\}$ such that $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers	(Q) False
(C) $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}$ equals	(R) 1
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List - I	List - II
(A) For each \(z_k\) there exists a \(z_j\) such \(z_k \cdot z_j=1\)	(P) True
(B) There exists a \(k \in\{1,2, \ldots, 9\}\) such that \(z_1 \cdot z=z_k\) has no solution z in the set of complex numbers	(Q) False
(C) \(\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}\) equals	(R) 1
(D) \(1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)\) equals	(S) 2

Let x∈R and let P=111022003, Q=2xx040xx6 and R=PQP-1.Then which of the following options is/are correct?

Step-by-step Solution

Let $x \in R$ and let $P = [\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix}], Q = [\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{matrix}]$ and $R = P Q P^{- 1} .$
Then which of the following options is/are correct?