Let \( A =
\begin{bmatrix}
1 & 0 & 0\\
3 & 1 & 0\\
9 & 3 & 1
\end{bmatrix}
\) and \( B = [b_{ij}], 1 \le i,j \le 3 \). If \( B = A^{99} - I \), then the value of \( \dfrac{b_{31}-b_{21}}{b_{32}} \) is:
- \(99\)
- \(199\)
- \(149\)
- \(159\)
The Correct Option is A
Solution and Explanation
Concept:
Matrices of the form \(A = I + N\), where \(N\) is nilpotent (\(N^k = 0\) for some \(k\)), can be simplified using the binomial theorem: \[ (I+N)^m = I + mN + \binom{m}{2}N^2 + \cdots \]
Step 1: Write the matrix in the form \(A = I + N\).
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix} = I + \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \] Let \[ N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \]
Step 2: Compute \(N^2\).
\[ N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} \] \[ N^3 = 0 \]
Step 3: Apply the binomial theorem.
\[ A^{99} = (I+N)^{99} \] \[ = I + 99N + \binom{99}{2}N^2 \] \[ \binom{99}{2} = 4851 \] \[ A^{99} = I + 99N + 4851N^2 \]
Step 4: Find matrix \(B\).
\[ B = A^{99} - I = 99N + 4851N^2 \] Compute required entries:
For \(b_{21}\): \[ b_{21} = 99 \times 3 = 297 \] For \(b_{32}\): \[ b_{32} = 99 \times 3 = 297 \] For \(b_{31}\): \[ b_{31} = 99 \times 9 + 4851 \times 9 \] \[ = 891 + 43659 = 44550 \]
Step 5: Compute the required value.
\[ \frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} \] \[ = \frac{44253}{297} = 149 \] Thus, the correct option is \(149\).
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