For x -0- define - A - begin bmatrix x-dfrac1x- 0 - 0 10 - x - 0 1080 - 16 lendbmatrix - B - beginbmatrix dfrac5 xx- 2-1- 0 - 0 l- dfrac3 x - 0 10 - 0 -14 e n d b m a t r i x -Let - X - AB -1- AB -2- cdots - AB - n 1Y - lim Vimitsn X Z - Y -1-2 ISwhere I is an identity matrix of order 3 - array - c - c -WlineWline

The correct option is C (C) → (R)A = [ x+1x amp;0 nbsp; amp;0; 0 amp;x nbsp; amp;0; 0 amp;0 nbsp; amp; 16; ] B = [ 5xx^ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Composite Function

For x >0, def...

Question

For $x > 0$ , define
$A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$ ,

$B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 x}{x^{2} + 1} & 0 & 0 0 & \frac{3}{x} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

Let $X = (A B)^{- 1} + (A B)^{- 2} + \dots + (A B)^{- n} Y = lim n \to \infty X Z = Y^{- 1} - 2 I$
where $I$ is an identity matrix of order $3$ .

$\begin{matrix} Column I & Column II (A) The minimum value of [trace (A Y)] is {[.] represents the greatest integer function} & (P) 24 (B) The value of det (Y^{- 1}) is & (Q) 12 (C) If trace (Z + Z^{2} + Z^{3} + \dots + Z^{10}) = 2^{a} + b (a, b \in N), then a + b is equal to & (R) 6 (D) If | adj (\sqrt{5} Y^{- 1}) | = k, then the number of odd positive divisors of k is & (S) 19 \end{matrix}$

Note: Trace of a square matrix is the sum of the diagonal elements.

Which of the following is the INCORRECT combination?

(C) \to (S)

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(D) \to (Q)

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(C) \to (R)

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(B) \to (P)

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Solution

The correct option is C $(C) \to (R)$
$A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$

$B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 x}{x^{2} + 1} & 0 & 0 0 & \frac{3}{x} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow A B = ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 0 0 & 3 & 0 0 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow (A B)^{- 1} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} & 0 & 0 0 & \frac{1}{3} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow (A B)^{- 2} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5^{2}} & 0 & 0 0 & \frac{1}{3^{2}} & 0 0 & 0 & \frac{1}{4^{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow (A B)^{- n} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5^{n}} & 0 & 0 0 & \frac{1}{3^{n}} & 0 0 & 0 & \frac{1}{4^{n}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$X = (A B)^{- 1} + (A B)^{- 2} + \dots + (A B)^{- n} \Rightarrow X = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} + \frac{1}{5^{2}} + \dots + \frac{1}{5^{n}} & 0 & 0 0 & \frac{1}{3} + \frac{1}{3^{2}} + \dots + \frac{1}{3^{n}} & 0 0 & 0 & \frac{1}{4} + \frac{1}{4^{2}} + \dots + \frac{1}{4^{n}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$Y = lim x \to \infty X \Rightarrow Y = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} & 0 & 0 0 & \frac{1}{2} & 0 0 & 0 & \frac{1}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow Y^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 0 0 & 2 & 0 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$

$Z = Y^{- 1} - 2 I \Rightarrow Z = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 0 0 & 0 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$(A)$
$A Y = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} & 0 & 0 0 & \frac{1}{2} & 0 0 & 0 & \frac{1}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow A Y = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{x^{2} + 1}{4 x} & 0 & 0 0 & \frac{x}{2} & 0 0 & 0 & \frac{16}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow trace (A Y) = \frac{x^{2} + 1}{4 x} + \frac{x}{2} + \frac{16}{3}$
Let $y = \frac{x^{2} + 1}{4 x} + \frac{x}{2} + \frac{16}{3}$
$y = \frac{3 x}{4} + \frac{1}{4 x} + \frac{16}{3} \Rightarrow y^{'} = \frac{3}{4} - \frac{1}{4 x^{2}} \Rightarrow y^{''} = \frac{2}{4 x^{3}}$

For minima,
$y^{'} = 0 \Rightarrow x = \pm \frac{1}{\sqrt{3}}$
Rejecting the negative value, so for $x = \frac{1}{\sqrt{3}}$
$y^{''} > 0$
Therefore, $x = \frac{1}{\sqrt{3}}$ is a minima point,
$y = \frac{16}{3} + \frac{\sqrt{3}}{2}$
The minimum of $[trace (A Y)] = 6$
$(A) \to (R)$

$(B)$
$Y^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 0 0 & 2 & 0 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow det (Y^{- 1}) = 24$
$(B) \to (P)$

$(C)$
$Z = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 0 0 & 0 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow Z^{2} = ⎡ ⎢ ⎣ \begin{matrix} 2^{2} & 0 & 0 0 & 0 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow Z^{10} = ⎡ ⎢ ⎣ \begin{matrix} 2^{10} & 0 & 0 0 & 0 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow trace (Z + Z^{2} + Z^{3} + \dots + Z^{10}) = (2 + 1) + (2^{2} + 1) + \dots + (2^{10} + 1) = \frac{2 (2^{10} - 1)}{2 - 1} + 10 = 2^{11} + 8 \Rightarrow 2^{a} + b = 2^{11} + 8 \Rightarrow a + b = 19$
$(C) \to (S)$

$(D)$
$Y^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 0 0 & 2 & 0 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow \sqrt{5} Y^{- 1} = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 4 \sqrt{5} & 0 & 0 0 & 2 \sqrt{5} & 0 0 & 0 & 3 \sqrt{5} \end{matrix} ⎤ ⎥ ⎥ ⎦$

We know that
$| adj Y | = | Y |^{n - 1} \Rightarrow | adj \sqrt{5} Y^{- 1} | = | \sqrt{5} Y^{- 1} |^{2} \Rightarrow | adj \sqrt{5} Y^{- 1} | = (4 \sqrt{5} \times 2 \sqrt{5} \times 3 \sqrt{5})^{2} \Rightarrow k = 2^{6} \times 3^{2} \times 5^{3}$

Number of odd divisors of $k$ ,
$= (2 + 1) (3 + 1) = 12$
$(D) \to (Q)$

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Similar questions

Q. For

x > 0,

let

A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

and

B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 x}{x^{2} + 1} & 0 & 0 0 & \frac{3}{x} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

be two matrices.
Three other matrices

X, Y

and

Z

are defined as

X = (A B)^{- 1} + (A B)^{- 2} + {A B}^{- 3} + \dots (A B)^{- n},

Y = lim n \to \infty X

and

Z = Y^{- 1} - 2 I,

where

I

is the identity matrix of order

3.

t r (P)

denotes the trace of matrix

P

and

a d j (P)

denotes the adjoint of matrix

P

, then which of the following is/are CORRECT?

If x,y, z are non-zero real numbers, then the inverse of matrix

$A = ⎡ ⎢ ⎣ \begin{matrix} x & 0 & 0 0 & y & 0 0 & 0 & z \end{matrix} ⎤ ⎥ ⎦$ is

a) $⎡ ⎢ ⎣ \begin{matrix} x^{- 1} & 0 & 0 0 & y^{- 1} & 0 0 & 0 & z^{- 1} \end{matrix} ⎤ ⎥ ⎦$

b) $x y z ⎡ ⎢ ⎣ \begin{matrix} x^{- 1} & 0 & 0 0 & y^{- 1} & 0 0 & 0 & z^{- 1} \end{matrix} ⎤ ⎥ ⎦$

c) $\frac{1}{x y z} ⎡ ⎢ ⎣ \begin{matrix} x & 0 & 0 0 & y & 0 0 & 0 & z \end{matrix} ⎤ ⎥ ⎦$

d) $\frac{1}{x y z} ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 0 & 1 & 1 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

. Then for positive integer

n, A^{n}

Given $l = ⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠$ Find $l^{2}$

Q. For

x > 0,

let

A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

and

B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 x}{x^{2} + 1} & 0 & 0 0 & \frac{3}{x} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

be two matrices.
Three other matrices

X, Y

and

Z

are defined as

X = (A B)^{- 1} + (A B)^{- 2} + {A B}^{- 3} + \dots (A B)^{- n},

Y = lim n \to \infty X

and

Z = Y^{- 1} - 2 I,

where

I

is the identity matrix of order

3.

t r (P)

denotes the trace of matrix

P

and

a d j (P)

denotes the adjoint of matrix

P

, then which of the following is/are CORRECT?

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