Calculating Matrix Elements for $B = A^{99} - I$
The problem asks for the value of $\frac{b_{31} - b_{21}}{b_{32}}$, where $B = A^{99} - I$ and $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$.
Matrix Power Calculation Strategy
We can express matrix $A$ as $A = I + N$, where $I$ is the identity matrix and $N = A - I$. $N = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$.
Let's find powers of $N$: $N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$. $N^3 = N^2 \cdot N = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Since $N^3 = 0$, $N$ is a nilpotent matrix. We can use the binomial theorem to calculate $A^{99}$: $A^{99} = (I+N)^{99} = \binom{99}{0}I^{99}N^0 + \binom{99}{1}I^{98}N^1 + \binom{99}{2}I^{97}N^2 + \binom{99}{3}I^{96}N^3 + \dots$
Because $N^3 = 0$, all terms from $N^3$ onwards are zero. $A^{99} = I + 99N + \frac{99(99-1)}{2}N^2$ $A^{99} = I + 99N + \frac{99 \times 98}{2}N^2$ $A^{99} = I + 99N + (99 \times 49)N^2$ $A^{99} = I + 99N + 4851N^2$.
Calculating Matrix B
Now, we find $B = A^{99} - I$. $B = (I + 99N + 4851N^2) - I$ $B = 99N + 4851N^2$.
Substitute the matrices $N$ and $N^2$: $B = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 4851 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ $B = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 & 297 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 43659 & 0 & 0 \end{bmatrix}$ $B = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 + 43659 & 297 & 0 \end{bmatrix}$ $B = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{bmatrix}$.
Finding the Required Value
From matrix $B$, we identify the elements: $b_{21} = 297$ $b_{31} = 44550$ $b_{32} = 297$
Calculate the required expression $\frac{b_{31} - b_{21}}{b_{32}}$: $\frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297}$ $\frac{44550 - 297}{297} = \frac{44253}{297}$
Performing the division: $\frac{44253}{297} = 149$.
Therefore, the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is 149.
Similar Questions
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Important Questions from Algebra
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
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