Question

Let a, b, c be such that b(a + c) ≠ 0 if

`|(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| + |(a + 1, b + 1, c - 1),(a - 1, b - 1, c + 1),((-1)^(n + 2)a, (-1)^(n + 1)b, (-1)^n c)|` = 0, then the value of n is ______.

Options

• any even integer

• any odd integer

• any integer

• zero

Answer 1

Let a, b, c be such that b(a + c) ≠ 0 if

`|(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| + |(a + 1, b + 1, c - 1),(a - 1, b - 1, c + 1),((-1)^(n + 2)a, (-1)^(n + 1)b, (-1)^n c)|` = 0, then the value of n is any odd integer.

Explanation:

`|(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| + |(a + 1, b + 1, c - 1),(a - 1, b - 1, c + 1),((-1)^(n + 2)a, (-1)^(n + 1)b, (-1)^n c)|` = 0

Taking transpose of the second determinant

`\implies |(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| + |(a + 1, a - 1, (-1)^(n + 2)a),(b + 1, b - 1, (-1)^(n + 1)b),(c - 1, c + 1, (-1)^n c)|` = 0

Apply C₁ `⇔` C₃ in second determinant

`\implies |(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| - |((-1)^(n + 2)a, a - 1, a + 1),((-1)^(n + 2)(-b), b - 1, b + 1),((-1)^(n + 2)c, c + 1, c - 1)|` = 0

Apply C₂ `⇔` C₃ in second determinant

`\implies |(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)| + (-1)^(n + 2) |(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)|` = 0

`\implies [1 + (-1)^(n + 2)] |(a, a + 1, a - 1),(-b, b + 1, b - 1),(c, c - 1, c + 1)|` = 0 

Apply C₂ `rightarrow` C₂ – C₁, C₃ `rightarrow` C₃ – C₁

`\implies [1 + (-1)^(n + 2)] |(a, 1, -1),(-b, 2b + 1, 2b - 1),(c, - 1, 1)|` = 0 

Apply R₁ `rightarrow` R₁ + R₃

`\implies [1 + (-1)^(n + 2)] |(a + c, 0, 0),(-b, 2b + 1, 2b - 1),(c, - 1, 1)|` = 0 

`\implies` [1 + (–1)^{n + 2}](a + c) (2b + 1 + 2b – 1) = 0

`\implies` 4b (a + c) [1 + (–1)^{n + 2}] = 0

`\implies` 1 + (–1)^{n + 2} = 0 as b (a + c) ≠ 0

`\implies` n should be an odd integer.