The value of \( x \) is obtained by solving the equation: \[ x^3 - 2x^2 - 9x + 18 = 0 \] Factoring this cubic equation gives: \[ (x - 2)(x^2 + 0x - 9) = 0 \] Solving for \( x \), we find \( x = 2 \). Substituting \( x = 2 \) into the matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & 8 & 9 \end{pmatrix} \] The determinant of \( A \) is given by: \[ \text{det}(A) = 1 \times \left| \begin{matrix} 2 & 6 \\ 8 & 9 \end{matrix} \right| - 2 \times \left| \begin{matrix} 4 & 6 \\ 7 & 9 \end{matrix} \right| + 3 \times \left| \begin{matrix} 4 & 2 \\ 7 & 8 \end{matrix} \right| \] Simplifying the determinants: \[ \text{det}(A) = 1 \times ((6 \times 8) - (2 \times 9)) - 2 \times ((4 \times 9) - (6 \times 7)) + 3 \times ((4 \times 8) - (2 \times 7)) \] \[ \text{det}(A) = 1 \times (18 - 48) - 2 \times (42-36) + 3 \times (32 - 14) \] \[ \text{det}(A) = 1 \times (30) - 2 \times (-6) + 3 \times 18 \] \[ \text{det}(A) = 30 + 12 + 54 = 96 \]
Therefore, the maximum value of \( A \) is 96.
First, we need to find the values of \( x \) that satisfy the equation \( x^3 - 2x^2 - 9x + 18 = 0 \). We can factor this equation:
\( x^2(x - 2) - 9(x - 2) = 0 \)
\( (x^2 - 9)(x - 2) = 0 \)
\( (x - 3)(x + 3)(x - 2) = 0 \)
So, the possible values of \( x \) are \( x = 3, x = -3, \) and \( x = 2 \).
Now, we need to find the determinant \( A \) for each value of \( x \):
\( A = \begin{vmatrix} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{vmatrix} \)
\( A = 1(9x - 48) - 2(36 - 42) + 3(32 - 7x) \)
\( A = 9x - 48 - 2(-6) + 96 - 21x \)
\( A = 9x - 48 + 12 + 96 - 21x \)
\( A = -12x + 60 \)
Now, we'll calculate \( A \) for each value of \( x \):
If \( x = 3 \): \( A = -12(3) + 60 = -36 + 60 = 24 \)
If \( x = -3 \): \( A = -12(-3) + 60 = 36 + 60 = 96 \)
If \( x = 2 \): \( A = -12(2) + 60 = -24 + 60 = 36 \)
The maximum value of \( A \) is 96.
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