If A=1/2 1 √3 -√3 1 then,

Correct answer: (C) A^30+A^25-A=I. Given A=1/2 1 √3 -√3 1 A^2=1/4 1 √3 -√3 1 1 √3 -√3 1 =1/4 1-3 2√3 -2√3 -3+1 A^2=1/2 -1 √3 -√3 -1 A^3=1/4 -1 √3 -√3 -1 -1 √3 -√3 -1 ⇒A^3=1/4 -4 0 0 -4 =- 1 0 0 1 ⇒A^3=-I .....(1) Now from equation 1, A^30=(-I)^10=I .....(2) Again from equation 1, A^25=A3^8.A=-I^8.A ⇒A^25=A ⇒A^25-A=0 ......(3) On adding equations 2 and 3 we get, A^30+A^25-A=I

Q46 / 90 · JEE Main 2023 (01 Feb Shift 2) · Mathematics

If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}]$ then,

A
$A^{30} - A^{25} = 2 I$
B
$A^{30} + A^{25} + A = I$
Correct answer
C
$A^{30} + A^{25} - A = I$
D
$A^{30} = A^{25}$

Solution

Given $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}]$

$A^{2} = \frac{1}{4} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}] [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}]$

$= \frac{1}{4} [\begin{matrix} 1 - 3 & 2 \sqrt{3} \\ - 2 \sqrt{3} & - 3 + 1 \end{matrix}]$

$A^{2} = \frac{1}{2} [\begin{matrix} - 1 & \sqrt{3} \\ - \sqrt{3} & - 1 \end{matrix}]$

$A^{3} = \frac{1}{4} [\begin{matrix} - 1 & \sqrt{3} \\ - \sqrt{3} & - 1 \end{matrix}] [\begin{matrix} - 1 & \sqrt{3} \\ - \sqrt{3} & - 1 \end{matrix}]$

$\Rightarrow A^{3} = \frac{1}{4} [\begin{matrix} - 4 & 0 \\ 0 & - 4 \end{matrix}] = - [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$\Rightarrow A^{3} = - I . . . . . (1)$

Now from equation $(1)$ ,

$A^{30} = (- I)^{10} = I . . . . . (2)$

Again from equation $(1)$ ,

$A^{25} = {(A^{3})}^{8} . A = {(- I)}^{8} . A$

$\Rightarrow A^{25} = A$

$\Rightarrow A^{25} - A = 0 . . . . . . (3)$

On adding equations $(2)$ and $(3)$ we get,

$A^{30} + A^{25} - A = I$

Hence, the correct answer is option: 'C'.

If A=1213-31 then,

Solution

Relevant Questions

If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}]$ then,