Let $\beta$ be a real number. Consider the matrix $$ A = \left(\begin{array}{ccc} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{array}\right) $$ If $A^7 - (\beta-1)A^6 - \beta A^5$ is a singular matrix, then the value of $9\beta$ is ____.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Matrices

Let

β

be a real number. Consider the matrix

A = β 23 011 1 - 2 - 2

A^{7} - (β - 1) A^{6} - β A^{5}

is a singular matrix, then the value of

9 β

is ____.

Answer: 3

First,

∣ A ∣ = - 1 \neq = 0

.

For

A^{7} - (β - 1) A^{6} - β A^{5}

to be singular:

∣ A^{7} - (β - 1) A^{6} - β A^{5} ∣ = 0

∣ A ∣^{5} ∣ A^{2} - (β - 1) A - β I ∣ = 0

Since

∣ A ∣^{5} = (- 1)^{5} = - 1 \neq = 0

, we need:

∣ A^{2} - β A - A + β I ∣ = 0

∣ A (A - β I) - I (A - β I) ∣ = 0

∣ (A - β I) (A + I) ∣ = 0

Compute

∣ A + I ∣

A + I = β + 1 23 021 1 - 2 - 1

∣ A + I ∣ = - 4 \neq = 0

Therefore,

∣ A - β I ∣ = 0

A - β I = 023 0 1 - β 1 1 - 2 - 2 - β

∣ A - β I ∣ = 2 - 3 (1 - β) = 3 β - 1 = 0

β = \frac{1}{3}

9 β = 3

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer