CameraIcon

SearchIcon

MyQuestionIcon

1

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

Multiplication of Matrices

For X >0, let...

Question

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 $3.$ If $t r (P)$ 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?

A

∣ ∣ a d j (\sqrt{5} Y^{- 1}) ∣ ∣ = 5 (5!)^{2}

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

B

∣ ∣ a d j (\sqrt{5} Y^{- 1}) ∣ ∣ = (5!)^{2}

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

C

The least positive integral value of

t r (A Y)

is

7

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

D

t r (Z + Z^{2} + Z^{3} + \dots + Z^{10}) = 2^{11} + 8

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

Open in App

Solution

The correct options are
A $∣ ∣ a d j (\sqrt{5} Y^{- 1}) ∣ ∣ = 5 (5!)^{2}$
C The least positive integral value of $t r (A Y)$ is $7$
D $t r (Z + Z^{2} + Z^{3} + \dots + Z^{10}) = 2^{11} + 8$
$A B = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{x^{2} + 1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 x}{x^{2} + 1} & 0 & 0 0 & \frac{3}{x} & 0 0 & 0 & \frac{1}{4} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow A B = ⎡ ⎢ ⎣ \begin{matrix} 5 & 0 & 0 0 & 3 & 0 0 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦$

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

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

$∴ X = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} + \frac{1}{5^{2}} + \dots + \frac{1}{5^{n}} & 0 & 0 0 & \frac{1}{3} + \frac{1}{3^{2}} + \dots + \frac{1}{3^{n}} & 0 0 & 0 & \frac{1}{4} + \frac{1}{4^{2}} + \dots + \frac{1}{4^{n}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Y = lim n \to \infty X = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{\frac{1}{5}}{1 - \frac{1}{5}} & 0 & 0 0 & \frac{\frac{1}{3}}{1 - \frac{1}{3}} & 0 0 & 0 & \frac{\frac{1}{4}}{1 - \frac{1}{4}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow Y = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} & 0 & 0 0 & \frac{1}{2} & 0 0 & 0 & \frac{1}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ∴ Y^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 0 0 & 2 & 0 0 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦$

$∣ ∣ a d j (\sqrt{5} Y^{- 1}) ∣ ∣ = ∣ ∣ (\sqrt{5})^{2} a d j (Y^{- 1}) ∣ ∣ = 5^{3} \cdot | Y^{- 1} |^{2} = 5^{3} 24^{2} = 5 \cdot (5!)^{2}$

$A Y = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} x + \frac{1}{x} & 0 & 0 0 & x & 0 0 & 0 & 16 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} & 0 & 0 0 & \frac{1}{2} & 0 0 & 0 & \frac{1}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow A Y = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} (x + \frac{1}{x}) & 0 & 0 0 & \frac{x}{2} & 0 0 & 0 & \frac{16}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \Rightarrow t r (A Y) = \frac{x}{4} + \frac{1}{4 x} + \frac{x}{2} + \frac{16}{3} \Rightarrow t r (A Y) = \frac{3 x}{4} + \frac{1}{4 x} + \frac{16}{3}$
As $x > 0$ , using $A . M . \geq G . M .$ , we get
$\frac{3 x}{4} + \frac{1}{4 x} \geq 2 \sqrt{\frac{3}{16}} \Rightarrow \frac{3 x}{4} + \frac{1}{4 x} \geq \frac{\sqrt{3}}{2}$

So, $t r (A Y) \geq \frac{\sqrt{3}}{2} + \frac{16}{3}$
$∴$ Least positive integral value of $t r (A Y) = 7$

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

$t r (Z + Z^{2} + Z^{3} + \dots + Z^{10}) = 2 + 2^{2} + 2^{3} + \dots + 2^{10} + 10 = 2 \cdot (\frac{2^{10} - 1}{2 - 1}) + 10 = 2^{11} + 8$

Suggest Corrections

thumbs-up

0

Similar questions

x > 0,

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} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

be two matrices.
Three other matrices

X, Y

Z

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

Y = lim n \to \infty X

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

I

is the identity matrix of order

3.

t r (P)

denotes the trace of matrix

P

a d j (P)

denotes the adjoint of matrix

P

, then which of the following is/are CORRECT?

x > 0,

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} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

be two matrices.
Two other matrices

X

Y

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

Y = lim n \to \infty X

t r (P)

denotes the trace of matrix

P

, then the least positive integral value of

t r (A Y)

x > 0

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} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

X = (A B)^{- 1} + (A B)^{- 2} + \dots + (A B)^{- n} Y = lim n \to \infty X Z = Y^{- 1} - 2 I

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?

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

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 7 0 & 1 & 0 1 & - 2 & 1 \end{matrix} ⎤ ⎥ ⎦

B = ⎡ ⎢ ⎣ \begin{matrix} - x & 14 x & 7 x 0 & 1 & 0 x & - 4 x & - 2 x \end{matrix} ⎤ ⎥ ⎦

be two matrices such that

A B = (A B)^{- 1}

A B \neq I,

I

is an identity matrix of order

3 \times 3.

Then the value of

t r (A B + (A B)^{2} + (A B)^{3} + \dots + (A B)^{100})

(

t r (A)

denotes the trace of matrix

A,

i.e., sum of diagonal elements of

A .)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Multiplication of Matrices

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Multiplication of Matrices

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon