<p>Let <em>a</em>, <em>b</em>, and <em>c</em> be such that <em>b</em>(<em>a</em> + <em>c</em>) ≠ 0. If</p><p>\[\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix} + \begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix} = 0,\] then the value of <em>n</em> is</p>
Step-by-Step Solution
Key Concept: Recognize that the first determinant can be simplified using column operations (C₂ - C₁ and C₃ - C₁), and the second determinant's rows can be reordered with sign changes to relate it to the first determinant's structure. The constraint that their sum equals zero determines n uniquely.
<p><strong>Step 1:</strong> Simplify the first determinant using column operations.</p><p>For the first determinant, apply C₂ → C₂ - C₁ and C₃ → C₃ - C₁:</p><p>$$\begin{vmatrix} a & 1 & -1 \\ -b & 1 & -1 \\ c & -1 & 1 \end{vmatrix}$$</p><p>Notice C₂ and C₃ are related: C₃ = -C₂. Therefore the first determinant = 0.</p><p><strong>Step 2:</strong> Analyze the second determinant condition.</p><p>Since the first determinant equals 0, the second determinant must also equal 0:</p><p>$$\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Determine n by examining linear dependence.</p><p>For this determinant to be zero, the third row must be a linear combination of the first two rows. Testing when Row₃ = -(Row₁ + Row₂):</p><p>Row₁ + Row₂ = (2a, 2b, 2c)</p><p>For Row₃ = -(2a, 2b, 2c) = (-2a, -2b, -2c):</p><p>We need: (-1)^(n+2) = -2, (-1)^(n+1) = -2, (-1)^n = -2</p><p>This is impossible. Instead, if Row₃ = -Row₁ + Row₂ = (-2a, -2b, 2c), we need (-1)^n = 1, so n must be even.</p><p><strong>Step 4:</strong> Verify with specific value.</p><p>Testing n = 2 (even): (-1)⁴ = 1, (-1)³ = -1, (-1)² = 1</p><p>The third row becomes: (-a, -b, c), which correctly makes the determinant zero through row operations.</p><p>∴ <strong>Answer: C (n = 2)</strong></p>
Correct Answer: C