Let x - R and let P- 1 1 1- 0 2 2- 0 0 3 - Q- 2 x x- 0 4 0- x x 6 - and R-PQP-1- Then which of the following options is-are correct-

Option 2: L.H.S. R=PQP^ 1 det R=|R|=|PQP^ 1|=|P||Q||P^ 1| =|P||Q|1|P|=|Q| ⇒ |R|=|Q|=48 4x^2 (∵det Q=|Q|=2 amp;x amp;x 0 amp;4 amp;0 x amp;x amp;6=4 ...

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Byju's Answer

Standard X

Mathematics

Identity Matrix

Let x ∈ R and...

Question

$Let x \in R$ 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?

For x = 1

, there exists a unit vector

α^i + β^j + γ^k for which R ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦

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det R = det ⎡ ⎢ ⎣ \begin{matrix} 2 & x & x 0 & 4 & 0 x & x & 5 \end{matrix} ⎤ ⎥ ⎦ + 8, for all x \in R

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There exists a real number

x

such that

P Q = Q P

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For x = 0, if R ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦ = 6 ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦,

then

a + b = 5

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Solution

The correct option is D $For x = 0, if R ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦ = 6 ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦,$ then $a + b = 5$
Option 2:
L.H.S.
$R = P Q P^{- 1}$
$det R = | R | = | P Q P^{- 1} | = | P | | Q | | P^{- 1} | = | P | | Q | \frac{1}{| P |} = | Q |$
$\Rightarrow | R | = | Q | = 48 - 4 x^{2}$
$⎛ ⎜ ⎝ ∵ det Q = | Q | = ∣ ∣ ∣ ∣ \begin{matrix} 2 & x & x 0 & 4 & 0 x & x & 6 \end{matrix} ∣ ∣ ∣ ∣ = 48 - 4 x^{2} ⎞ ⎟ ⎠$
R.H.S.
$det ⎡ ⎢ ⎣ \begin{matrix} 2 & x & x 0 & 4 & 0 x & x & 5 \end{matrix} ⎤ ⎥ ⎦ + 8 = 40 - 4 x^{2} + 8 = 48 - 4 x^{2} = det R$

Option 1:
$If P Q = Q P \Rightarrow P Q P^{- 1} = Q P P^{- 1} \Rightarrow R = Q$
$∵ P^{- 1} = \frac{1}{6} ⎡ ⎢ ⎣ \begin{matrix} 6 & - 3 & 0 0 & 3 & - 2 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦$ and
$R = P Q P^{- 1} = \frac{1}{6} ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 0 & 2 & 2 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 2 & x & x 0 & 4 & 0 x & x & 6 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 6 & - 3 & 0 0 & 3 & - 2 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦$

$= \frac{1}{6} ⎡ ⎢ ⎣ \begin{matrix} 6 x + 12 & 3 x + 6 & 4 - 10 x 12 x & 24 & 8 - 4 x 18 x & 0 & 36 - 6 x \end{matrix} ⎤ ⎥ ⎦$
$Put x = 0$
$R = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 2 & 1 & \frac{2}{3} 0 & 4 & \frac{4}{3} 0 & 0 & 6 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
For any value of $x, R \neq Q$
Therefore, $P Q = Q P$ doesn't exist for any real number $x .$

Option 3:
For $x = 0$ $R ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦ = 6 ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 2 & 1 & \frac{2}{3} 0 & 4 & \frac{4}{3} 0 & 0 & 6 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 a b \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 6 6 a 6 b \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 2 + a + \frac{2 b}{3} 4 a + \frac{4 b}{3} 6 b \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 6 6 a 6 b \end{matrix} ⎤ ⎥ ⎦$
Comparing the element of both the matrix, we get
$2 + a + \frac{2 b}{3} = 6 . . . (i) and 4 a + \frac{4 b}{3} = 6 a . . . (i i)$
Solving both the equation
$a = 2, b = 3 \Rightarrow a + b = 5$

Option 4:
$For x = 1, R = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 3 & \frac{3}{2} & - 1 2 & 4 & \frac{2}{3} 3 & 0 & 5 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$R ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 3 α + \frac{3}{2} β - γ 2 α + 4 β + \frac{2}{3} γ 3 α + 5 γ \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦$
After solving, we will get trivial soloution i.e.,
$α = β = γ = 0$
So for $x = 1,$ there is no unit vector $α^i + β^j + γ^k for which R ⎡ ⎢ ⎣ \begin{matrix} α β γ \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦$

Suggest Corrections

Let x∈R and let P=⎡⎢⎣111022003⎤⎥⎦, Q=⎡⎢⎣2xx040xx6⎤⎥⎦ and R=PQP−1. Then which of the following options is/are correct?

$Let x \in R$ 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?