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 \( a_{22} = 2 \).
\( 12 \)
We are given a \( 3 \times 3 \) matrix \( A = [a_{ij}] \), where each row sum is equal to 6 and \( a_{22} = 2 \).
Step 1: Construct the Matrix Using Given Conditions
From the row sum condition: \[ a_{11} + a_{12} + a_{13} = 6 \] \[ a_{21} + a_{22} + a_{23} = 6 \] \[ a_{31} + a_{32} + a_{33} = 6 \] Using the given condition that for \( i<3 \), we have: \[ a_{ii} = a_{ij} + a_{in}, \quad j = i + 1 \] For \( i = 1 \): \[ a_{11} = a_{12} + a_{13} \] For \( i = 2 \): \[ a_{22} = a_{23} + a_{21} \] Since \( a_{22} = 2 \), we substitute: \[ 2 = a_{23} + a_{21} \] For \( i = 3 \), we use: \[ a_{33} = a_{31} + a_{32}, \quad j = 4 - i \]
Step 2: Solve for Matrix Elements
From row sums: \[ a_{11} + a_{12} + a_{13} = 6 \] Using \( a_{11} = a_{12} + a_{13} \), we get: \[ (a_{12} + a_{13}) + a_{12} + a_{13} = 6 \] \[ 2(a_{12} + a_{13}) = 6 \] \[ a_{12} + a_{13} = 3, \quad a_{11} = 3 \] Similarly, using \( a_{22} = 2 \): \[ a_{21} + a_{23} = 2, \quad a_{21} + 2 + a_{23} = 6 \] \[ a_{21} + a_{23} = 4 \] Solving, we find: \[ a_{21} = 1, \quad a_{23} = 1 \] For \( i = 3 \): \[ a_{31} + a_{32} + a_{33} = 6 \] Using \( a_{33} = a_{31} + a_{32} \), we substitute: \[ (a_{31} + a_{32}) + a_{31} + a_{32} = 6 \] \[ 2(a_{31} + a_{32}) = 6 \] \[ a_{31} + a_{32} = 3, \quad a_{33} = 3 \]
Step 3: Compute Determinant
The determinant of a \( 3 \times 3 \) matrix is given by: \[ |A| = a_{11} (a_{22} a_{33} - a_{23} a_{32}) - a_{12} (a_{21} a_{33} - a_{23} a_{31}) + a_{13} (a_{21} a_{32} - a_{22} a_{31}) \] \[ = 3(2 \cdot 3 - 1 \cdot 1) - 1(1 \cdot 3 - 1 \cdot 2) + 2(1 \cdot 1 - 2 \cdot 2) \] \[ = 3(6 - 1) - 1(3 - 2) + 2(1 - 4) \] \[ = 3(5) - 1(1) + 2(-3) \] \[ = 15 - 1 - 6 \] \[ = 12 \] Thus, the determinant \( |A| \) is \( 12 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]
\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} \]
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
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 \( 11(a + b + c) \)?
\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]