If \( a \neq b \neq c \), then
\[ \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \]and
\[ \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \]then what is \( 11(a + b + c) \)?
We are given the determinants \( \Delta_1 \) and \( \Delta_2 \), as well as the ratio \( \frac{\Delta_1}{\Delta_2} \). Our goal is to compute \( 11(a + b + c) \).
Step 1: We begin by simplifying the determinant \( \Delta_1 \). First, evaluate \( \Delta_1 \):
\[ \Delta_1 = \begin{vmatrix} 1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix} \]We expand this determinant along the first row:
\[ \Delta_1 = 1 \times \begin{vmatrix} b^2 & ca \\ c^2 & ab \end{vmatrix} - a^2 \times \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} + bc \times \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} \]Now calculate the 2×2 determinants:
\[ \begin{vmatrix} b^2 & ca \\ c^2 & ab \end{vmatrix} = b^2 \cdot ab - ca \cdot c^2 = ab^3 - ac^3 \] \[ \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} = 1 \cdot ab - ca \cdot 1 = ab - ac \] \[ \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1 \cdot c^2 - b^2 \cdot 1 = c^2 - b^2 \]Substituting these values back into the expression for \( \Delta_1 \):
\[ \Delta_1 = ab^3 - ac^3 - a^2 (ab - ac) + bc (c^2 - b^2) \] \[ \Delta_1 = ab^3 - ac^3 - a^3b + a^3c + bc^3 - bc^2b \] \[ \Delta_1 = ab^3 - ac^3 - a^3b + a^3c + bc^3 - b^3c \]Step 2: Next, simplify the determinant \( \Delta_2 \):
\[ \Delta_2 = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \]Expanding along the first row:
\[ \Delta_2 = 1 \times \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} - 1 \times \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} + 1 \times \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} \]Now calculate the 2×2 determinants:
\[ \begin{vmatrix} b^2 & c^2 \\ b^3 & c^3 \end{vmatrix} = b^2c^3 - b^3c^2 \] \[ \begin{vmatrix} a^2 & c^2 \\ a^3 & c^3 \end{vmatrix} = a^2c^3 - a^3c^2 \] \[ \begin{vmatrix} a^2 & b^2 \\ a^3 & b^3 \end{vmatrix} = a^2b^3 - a^3b^2 \]Substituting these into \( \Delta_2 \):
\[ \Delta_2 = b^2c^3 - b^3c^2 - a^2c^3 + a^3c^2 + a^2b^3 - a^3b^2 \]Step 3: Given the ratio \( \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \), we can now solve for \( 11(a + b + c) \).
By properties of determinants, the solution simplifies to:
\[ 11(a + b + c) = 6(ab + bc + ca) \]
Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]
\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} \]
The roots of the equation \( x^3 - 3x^2 + 3x + 7 = 0 \) are \( \alpha, \beta, \gamma \) and \( w, w^2 \) are complex cube roots of unity. If the terms containing \( x^2 \) and \( x \) are missing in the transformed equation when each one of these roots is decreased by \( h \), then
\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]