The value of \( x \) that satisfies the equation:
\[ \begin{vmatrix} x & 1 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & -2 \end{vmatrix} = 6 \]
To find the value of \( x \) in a determinant equation, first expand the determinant of the given matrix. Use the formula for the determinant of a \(3 \times 3\) matrix:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
After expanding, solve for \( x \) by setting the result equal to the given value.
We are given the determinant equation:
\[ \begin{vmatrix} x & 1 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & -2 \end{vmatrix} = 6 \]
We need to evaluate the determinant of this matrix.
The determinant of a \(3 \times 3\) matrix is calculated using the formula:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For the given matrix:
\[ \begin{vmatrix} x & 1 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & -2 \end{vmatrix} \]
We have \( a = x \), \( b = 1 \), \( c = 1 \), \( d = 2 \), \( e = 2 \), \( f = 0 \), \( g = 1 \), \( h = 0 \), and \( i = -2 \). Applying the determinant formula:
\[ = x \left( 2(-2) - 0(0) \right) - 1 \left( 2(-2) - 0(1) \right) + 1 \left( 2(0) - 2(1) \right) \]
\[ = x(-4) - 1(-4) + 1(-2) \]
\[ = -4x + 4 - 2 \]
\[ = -4x + 2 \]
Setting this equal to 6:
\[ -4x + 2 = 6 \]
\[ -4x = 4 \]
\[ x = -1 \]
Thus, the value of \( x \) is \( -1 \).
Therefore, the correct answer is option (E), \( x = -1 \).
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: