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
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
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If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
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Important Questions from Algebra
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
