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.
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: