Question

If x³ - 2x² - 9x + 18 = 0 and A = \(\begin{vmatrix}     1 & 2 & 3 \\     4 & x & 6 \\     7 & 8 & 9 \\ \end{vmatrix}\) then the maximum value of A is

• A. 96
• B. 36
• C. 24
• D. 120

Answer 1

The Correct Option is A

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.