If
and \( AA^T = I \), then \( \frac{a}{b} + \frac{b}{a} = \):
Step 1: Solving for matrix \( A \). Given that \( 3A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \), we solve for \( A \): \[ A = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}. \] Step 2: Use the condition \( AA^T = I \). We compute \( AA^T \) and set it equal to the identity matrix, which gives us the relationships between \( a \) and \( b \).
Step 3: Solving for \( a \) and \( b \). From the equations, we find that \( a = -5 \) and \( b = 5 \).
Step 4: Compute \( \frac{a}{b} + \frac{b}{a} \). \[ \frac{a}{b} + \frac{b}{a} = \frac{-5}{5} + \frac{5}{-5} = -1 + (-1) = -2. \]
If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:
\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]
holds for: