Let $\mathrm{A}=\left[\begin{array}{lll}0 & 1 & 2 \\ \mathrm{a} & 0 & 3 \\ 1 & \mathrm{c} & 0\end{array}\right]$, where $\mathrm{a}, \mathrm{c} \in \mathrm{R}$. If $\mathrm{A}^{3}=\mathrm{A}$ and the positive value of a belongs to the interval $(\mathrm{n}-1, \mathrm{n}]$, where $\mathrm{n} \in \mathrm{N}$, then n is equal to

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Matrices

Let

A = 0 a 1 10 c 230

, where

a, c \in R

. If

A^{3} = A

and the positive value of a belongs to the interval

(n - 1, n]

, where

n \in N

, then n is equal to

Answer: 2

Given

A^{3} = A

, we compute

A^{2}

and

A^{3}

and equate corresponding entries.

A^{2} = a + 2 3 a c 2 c a + 3 c 1 3 2 a 2 + 3 c

Computing

A^{3} = A^{2} \cdot A

and using

A^{3} = A

, we compare the (1,1) entries:

2 a c + 3 = 0

... (1)

Comparing the (1,2) entries:

a + 2 + 3 c = 1

a + 3 c = - 1

... (2)

From equation (1):

c = - \frac{3}{2 a}

Substituting into equation (2):

a + 3 (- \frac{3}{2 a}) = - 1

a - \frac{9}{2 a} = - 1

2 a^{2} - 9 = - 2 a

2 a^{2} + 2 a - 9 = 0

Using the quadratic formula or checking sign changes:

The positive root lies in

(1, 2]

, so

n = 2

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer