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), \( x = -1 \).
The focus of the parabola \(y^2 + 4y - 8x + 20 = 0\) is at the point:
Let \( S \) denote the set of all subsets of integers containing more than two numbers. A relation \( R \) on \( S \) is defined by:
\[ R = \{ (A, B) : \text{the sets } A \text{ and } B \text{ have at least two numbers in common} \}. \]
Then the relation \( R \) is:
The centre of the hyperbola \(16x^2 - 4y^2 + 64x - 24y - 36 = 0\) is at the point: