Matrices & Determinants
Determinants
Grade 12
Question:
<p>If \(f(\theta) = \begin{vmatrix} 1 & \cos\theta & 1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}\begin{vmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{vmatrix}\) and \(A\) and \(B\) are respectively the maximum and the minimum values of \(f(\theta)\), then \((A, B)\) is equal to</p>
<p>\((3, -1)\)</p>
<p>\((4, 2-\sqrt{2})\)</p>
<p>\((2+\sqrt{2}, 2-\sqrt{2})\)</p>
<p>\((2+\sqrt{2}, -1)\)</p>
Step-by-Step Solution
Key Concept: First evaluate the 3×3 determinant as a function of θ using cofactor expansion, then recognize it's a quadratic in sin θ and cos θ. The second determinant is incomplete (2×3 matrix), so f(θ) likely equals just the first determinant; find its maximum and minimum by calculus or by recognizing its form.
<p><strong>Step 1:</strong> Expand the 3×3 determinant using the first row:</p><p>f(θ) = 1·|1 -cos θ| − cos θ·|-sin θ -cos θ| + 1·|-sin θ 1|</p><p> |sin θ 1 | |-1 1 | |-1 sin θ|</p><p><strong>Step 2:</strong> Compute each 2×2 minor:</p><p>• First: (1)(1) − (-cos θ)(sin θ) = 1 + sin θ cos θ</p><p>• Second: (-sin θ)(1) − (-cos θ)(-1) = -sin θ − cos θ</p><p>• Third: (-sin θ)(sin θ) − (1)(-1) = -sin² θ + 1 = cos² θ</p><p><strong>Step 3:</strong> Combine:</p><p>f(θ) = (1 + sin θ cos θ) − cos θ(-sin θ − cos θ) + cos² θ</p><p>= 1 + sin θ cos θ + sin θ cos θ + cos² θ + cos² θ</p><p>= 1 + 2 sin θ cos θ + 2cos² θ</p><p>= 1 + sin 2θ + (1 + cos 2θ)</p><p>= 2 + sin 2θ + cos 2θ</p><p><strong>Step 4:</strong> Express sin 2θ + cos 2θ = √2 sin(2θ + π/4)</p><p>Maximum of f(θ) = 2 + √2, Minimum of f(θ) = 2 − √2</p><p>∴ (A, B) = (2 + √2, 2 − √2)</p>
Correct Answer: C