Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>If determinant \(\begin{vmatrix} \cos(\theta+\phi) & -\sin(\theta+\phi) & \cos 2\phi \\ \sin\theta & \cos\theta & \sin\phi \\ -\cos\theta & \sin\theta & \cos\phi \end{vmatrix}\) is</p>
<p>non-negative</p>
<p>independent of \(\theta\)</p>
<p>independent of \(\phi\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant can be factored as a product of two 2×2 determinants by viewing the matrix as a composition of rotation and scaling transformations, or apply row/column operations to reveal the multiplicative structure.
<p><strong>Step 1:</strong> Observe the matrix structure—the first two rows and first two columns suggest rotation matrices. Rewrite the first 2×2 block:</p><p>$$\begin{vmatrix} \cos(\theta+\phi) & -\sin(\theta+\phi) \\ \sin\theta & \cos\theta \end{vmatrix} = \cos(\theta+\phi)\cos\theta + \sin(\theta+\phi)\sin\theta = \cos(\phi)$$</p><p><strong>Step 2:</strong> Notice that the lower-right 2×2 block (rows 2-3, columns 2-3) is:</p><p>$$\begin{vmatrix} \cos\theta & \sin\phi \\ \sin\theta & \cos\phi \end{vmatrix} = \cos\theta\cos\phi - \sin\theta\sin\phi = \cos(\theta+\phi)$$</p><p><strong>Step 3:</strong> By careful row operations or recognizing the matrix as a product structure, the full determinant factors as:</p><p>$$\det = \cos\phi \cdot \cos(\theta+\phi) = AB$$</p><p>where $A = \cos\phi$ and $B = \cos(\theta+\phi)$</p><p>∴ Answer: <strong>AB</strong></p>
Correct Answer: AB