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😃: A6 Let c_0, c_1, c_2, \dots be the sequence defined so that
for sufficiently small x. For a positive integer n, let A be the n-by-n matrix with i,j-entry c_{i+j-1} for i and j in {1, \dots, n}. Find the determinant of A.
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Okay, the sequence is defined recursively as ( \frac{1 - 3x - \sqrt{1 - 14x + 9x^2}}{4} = \sum_{k=0}^{\infty} c_k x^k ), where ( c_0 = 1 ) and ( c_1 = 3 ) and ( c_2 = \frac{1}{9} \sqrt{1 - 14 \cdot 3 + 9 \cdot 3^2} ).
Let's find the determinant of matrix A:
A = \begin{bmatrix}
1 - 3x & 3 & \frac{1}{9} \sqrt{1 - 42 + 81} \
3 & 1 - 3x & \frac{1}{9} \sqrt{1 - 42 + 81} \
0 & 0 & 1 - 3x
\end{bmatrix}
First, simplify the expression under the square root:
1 - 42 + 81 = 30,
\therefore \frac{1}{9} \sqrt{30} = \frac{1}{9} \cdot 5\sqrt{30} = \frac{5\sqrt{30}}{9}.
Now, let's calculate the determinant of matrix A:
\begin{align*}
\det(A) &= \begin{vmatrix}
1 - 3x & 3 & \frac{5\sqrt{30}}{9} \
3 & 1 - 3x & \frac{5\sqrt{30}}{9} \
0 & 0 & 1 - 3x
\end{vmatrix}
= (1-3x)\left[(1-3x)\left(\frac{5\sqrt{30}}{9}\right) - 3\cdot 0\right]
\quad \text{(expanding the determinant along the first row)}
= (1-3x)\left(\frac{5\sqrt{30}}{9} - 0\right)
= \frac{5\sqrt{30}}{9} (1-3x)
\end{align*}
We need to find the value of x such that (\frac{5\sqrt{30}}{9} (1-3x) = 0), which implies (1-3x = 0), so (3x = 1), and finally (x = \frac{1}{3}).
Therefore, the determinant of A is (5\sqrt{30} \cdot \frac{1}{3} = \frac{5\sqrt{30}}{3}).