Question

If X = $\left[\begin{array}{ccc}0&1\\0&0\end{array}\right]$ and (pI + qX)ˣ = AI + BX, ∀ p, q ∈ R, where A and B are scalars and I is two rowed unit matrix, then A and B are
(a) pˣ⁻¹, xpˣq
(b) pˣ, xpˣ⁻¹q
(c) pˣ, xpˣ⁻²q
(d) None of these

Answer 1

HEYA MATE HERE UR ANSWER,

The answer is in the given attachment

So the answer is option c.) p^x, xp^x-2q

Hope it helps u...

No need to mark as the brainliest...

Answer 2

Given:

⟼ A matrix

$\rm{X=\left[\begin{array}{cc}0&1\\0&0\end{array}\right]}$

⟼ (pI + qX)ˣ = AI + BX

To Find:

Value of A and B

Solution:

We know that,

$\longrightarrow\bf\pink{I^{n}=I}$

$\longrightarrow\bf\purple{AI=IA=A}$

where,

I is an identity matrix

A is any matrix

Order of I and A is same

$\rule{190}{1}$

Firstly, we have to find X²

$\longrightarrow\rm{X^{2}=X.X}$

$\longrightarrow\rm{X^{2}=\left[\begin{array}{cc}0&1\\0&0\end{array}\right]\left[\begin{array}{cc}0&1\\0&0\end{array}\right]}$

$\longrightarrow\rm{X^{2}=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]}$

Now, we have to find X³

$\longrightarrow\rm{X^{3}=X^{2}.X}$

$\longrightarrow\rm{X^{3}=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\left[\begin{array}{cc}0&1\\0&0\end{array}\right]}$

$\longrightarrow\rm{X^{3}=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]}$

Now, we will find X⁴

$\longrightarrow\rm{X^{4}=X^{3}.X}$

$\longrightarrow\rm{X^{4}=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\left[\begin{array}{cc}0&1\\0&0\end{array}\right]}$

$\longrightarrow\rm{X^{4}=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]}$

Just like this,

X² = X³ = X⁴ = X⁵ = ..........= Xˣ = O

where, x∈N and O is a null matrix

Now,

By using Binomial distribution, we get

$\longrightarrow\rm{(pI + qX)^{x}}$

$\sf{(pI)^{x}+{}^{x}C_{1}(pI)^{x-1}(qX)+{}^{x}C_{2}(pI)^{x-2}(qX)^{2}+....+(qX)^{x}}$

$\sf{p^{x}I^{x}+xp^{x-1}I^{x-1}(qX)+{}^{x}C_{2}(p^{x-2}I^{x-2})q^{2}X^{2}+....+q^{x}X^{x}}$

$\sf{p^{x}I+xp^{x-1}I(qX)+{}^{x}C_{2}(p^{x-2}I)q^{2}(O)+....+q^{x}(O)}$

$\rm{p^{x}I+xp^{x-1}I(qX)+0+....+0}$

$\rm{p^{x}I+xp^{x-1}IqX}$

$\rm{p^{x}I+xp^{x-1}q(IX)}$

$\rm{p^{x}I+xp^{x-1}qX}$

$\rm{(p^{x})I+(xp^{x-1}q)X}$

So,

$\rm{(pI + qX)^{x}=(p^{x})I+(xp^{x-1}q)X}---(1)$

But we know that,

(pI + qX)ˣ = AI + BX --------(2)

On comparing (1) and (2), we get

$\longrightarrow\rm\green{A=p^{x}}$

$\longrightarrow\rm\green{B=xp^{x-1}q}$

Hence, the correct option is option (b)