Matrices & Determinants
Properties of determinants
Grade 12
Question:
<p>If \(f(\alpha, \beta) = \begin{vmatrix} \cos\alpha & -\sin\alpha & 1 \\ \sin\alpha & \cos\alpha & 1 \\ \cos(\alpha+\beta) & -\sin(\alpha+\beta) & 1 \end{vmatrix}\), then</p>
<p>\(f(300, 200) = f(400, 200)\)</p>
<p>\(f(200, 400) = f(200, 600)\)</p>
<p>\(f(100, 200) = f(200, 200)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that the first two rows form a rotation matrix pattern. Use row operations to simplify the determinant by expressing the third row in terms of the first two rows using angle addition formulas.
<p><strong>Step 1:</strong> Observe the structure. Rows 1 and 2 represent a rotation matrix with angle α. Row 3 has the same structure with angle (α+β).</p><p><strong>Step 2:</strong> Use the identity: cos(α+β) = cos α cos β - sin α sin β and -sin(α+β) = -sin α cos β - cos α sin β.</p><p><strong>Step 3:</strong> Note that Row 3 = cos β · Row 1 - sin β · Row 2. This means Row 3 is a linear combination of Rows 1 and 2.</p><p><strong>Step 4:</strong> Since one row is a linear combination of other rows, the determinant equals <strong>0</strong> for all values of α and β.</p><p><strong>Step 5:</strong> Therefore f(α,β) = 0 regardless of α and β values, making statements about f(α,β) being constant (= 0) and independent of both α and β true.</p><p>∴ Answer: A,B (assuming these options state f(α,β) is constant and equals 0)</p>
Correct Answer: A,B