JEE Advanced 2022 — Paper 2 — MATH-I solution

JEE Advanced2022Paper 2MATH-I

Let $β$ be a real number. Consider the matrix $A = β 01 21 - 2 31 - 2$ . If $A^{7} - (β - 1) A^{6} - β A^{5}$ is a singular matrix, then the value of $9 β$ is _____.

Your answer

Solution

Factor: $A^{7} - (β - 1) A^{6} - β A^{5} = A^{5} [A^{2} - (β - 1) A - β I] = A^{5} (A + I) (A - β I)$ .

Compute $det (A)$ by expansion along the first row: $det (A) = β (1 \cdot (- 2) - (- 2) \cdot 1) - 0 + 1 \cdot (2 - 3) = β (0) + (- 1) = - 1$ .

So $det (A^{5}) = (- 1)^{5} = - 1 \neq = 0$ . The product is singular when $det (A + I) = 0$ or $det (A - β I) = 0$ .

$det (A + I) = β + 101 22 - 2 31 - 1 = (β + 1) (- 2 + 2) - 0 + 1 (2 - 6) = - 4 \neq = 0$ .

$det (A - β I) = 001 21 - β - 2 31 - 2 - β$ . Expanding along the first row: $1 \cdot (2 \cdot 1 - 3 (1 - β)) = 2 - 3 + 3 β = 3 β - 1$ .

Setting $3 β - 1 = 0$ gives $β = \frac{1}{3}$ , so $9 β = 3$ .

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