Let and , . If , then the value of is : [2026]

Question

Let $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$ and $B = [b_{i j}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is : [2026]

Smart Achievers · Accepted Answer

(3)

Given, $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$

$\Rightarrow A = I + N$ , where

$N = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}]$ and $I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow N^{2} = [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]$

$\Rightarrow N^{2} = N^{2} \cdot N = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$\Rightarrow A^{99} = {(I + N)}^{99} = I + 99 N + \frac{9998}{2} N^{2} + . . .$

Since, $N^{k} = 0$ , for $k \geq 3$

$∴ A^{99} = I + 99 N + 4851 N^{2}$

$B = I + 99 N + 4851 N^{2}$

$B = 99 N + 4851 N^{2}$

$= [\begin{matrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{matrix}] + 4851 [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 + 43659 & 297 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{matrix}]$

$∴ \frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = 149$

Solution

Q.
Let $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{matrix}]$ and $B = [b_{i j}]$ , $1 \leq i, j \leq 3$ . If $B = A^{99} - I$ , then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is : [2026]

1 99

2 199

3 149

4 159

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