If P- 5 -3- 111 336 - and -3P2013-P2014-2 1-2 where - are natural numbers- then

|P| = nbsp;5 amp; 3 111 amp;336 ⇒ |P| =1680+333 ⇒ |P|= 2013 Now, ( 3P^2013+P^2014) =|P^2013( 3I + P)| =|P|^ 2013 |P 3I| = (2013)^2013 nbsp;2 amp; 3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Rank of a Matrix

If P=[ 5 -3;...

Question

If $P = [\begin{matrix} 5 & - 3 111 & 336 \end{matrix}]$ and $det (- 3 P^{2013} + P^{2014}) = α^{α} β^{2} (1 + γ + γ^{2})$ where $α, β, γ$ are natural numbers, then

α = 2013

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

β = 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

γ = 10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

α = 2014

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $γ = 10$
$| P | = ∣ ∣ ∣ \begin{matrix} 5 & - 3 111 & 336 \end{matrix} ∣ ∣ ∣$
$\Rightarrow | P | = 1680 + 333$
$\Rightarrow | P | = 2013$

Now, $det (- 3 P^{2013} + P^{2014})$
$= ∣ ∣ P^{2013} (- 3 I + P) ∣ ∣$
$= | P |^{2013} | P - 3 I |$
$= (2013)^{2013} ∣ ∣ ∣ \begin{matrix} 2 & - 3 111 & 333 \end{matrix} ∣ ∣ ∣$
$= (2013)^{2013} \cdot (111 \times 9)$
$= (2013)^{2013} \cdot 3^{2} (1 + 10 + 10^{2})$
On comparing, we get
$α = 2013,$ $β = 3$ and $γ = 10$

Suggest Corrections

Similar questions

Q. If

P = [\begin{matrix} 5 & - 3 111 & 336 \end{matrix}]

and

det (- 3 P^{2013} + P^{2014}) = α^{α} β^{2} (1 + γ + γ^{2})

where

α, β, γ

are natural numbers, then

Q. Assertion :The minimum value of the expression

sin α + sin β + sin γ

where

α, β, γ

are real numbers such that

α + β + γ = π

is negative. Reason: If

α + β + γ = π

,then

α, β, γ

are the angles of a triangle.

Q. The points

(α, β), (γ, δ) (α, δ)

and

(γ, β)

taken in order, where

α, β, γ, δ

are different real numbers, are

Q. If

sin (2 α - β - γ) + 8 sin (2 β - γ - α) + 27 sin (2 γ - α - β) = P

and

sin (α + β) + sin (β + γ) + sin (γ + α) = Q

then P+Q is equal to

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

, then the value of determinant

Δ

is , where

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ β & γ & α γ & α & β \end{matrix} ∣ ∣ ∣ ∣ ∣

Join BYJU'S Learning Program

If P=[5−3111336] and det(−3P2013+P2014)=ααβ2(1+γ+γ2) where α,β,γ are natural numbers, then

The correct option is C γ=10|P|=∣∣∣5−3111336∣∣∣ ⇒|P|=1680+333 ⇒|P|=2013 Now, det(−3P2013+P2014) =∣∣P2013(−3I+P)∣∣ =|P|2013|P−3I| =(2013)2013∣∣∣2−3111333∣∣∣ =(2013)2013⋅(111×9) =(2013)2013⋅32(1+10+102) On comparing, we get α=2013, β=3 and γ=10

If $P = [\begin{matrix} 5 & - 3 111 & 336 \end{matrix}]$ and $det (- 3 P^{2013} + P^{2014}) = α^{α} β^{2} (1 + γ + γ^{2})$ where $α, β, γ$ are natural numbers, then