Let A be a matrix such that for all nonzero . If , and , then is __________. [2025]

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Solution

Q.
Let A be a $3 \times 3$ matrix such that $X^{T} A X =' O'$ for all nonzero $3 \times 1 matrices X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ . If $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}], A [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 4 \\ - 8 \end{matrix}]$ , and $\det (adj (2 (A + I))) = 2^{α} 3^{β} 5^{γ}, α, β, γ \in ℕ$ , then $α^{2} + β^{2} + γ^{2}$ is __________. [2025]

Ans.
(44)

Given, $X^{T} A X = O$

$∴ [\begin{matrix} X & Y & Z \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] [\begin{matrix} X \\ Y \\ Z \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}], where A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]$

$\Rightarrow X (a_{1} X + a_{2} Y + a_{3} Z) + Y (b_{1} X + b_{2} Y + b_{3} Z) + Z (c_{1} X + c_{2} Y + c_{3} Z) = 0$

On comparing cofficients, we get

$\Rightarrow a_{1} = 0, b_{2} = 0, c_{3} = 0 and a_{2} + b_{1} = 0, a_{3} + c_{1} = 0, b_{3} + c_{2} = 0$

$∴ A = [\begin{matrix} 0 & a_{2} & a_{3} \\ - a_{2} & 0 & b_{3} \\ - a_{3} & - b_{3} & 0 \end{matrix}] = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}] (Let)$

$∴ A = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}]$ , which is skew-symmetric matrix

Given, $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}] \Rightarrow A = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}]$

x + y = 1, – x + z = 4, y + z = – 5

$[\begin{matrix} 0 & x & y \\ - z & 0 & z \\ - y & - x & 0 \end{matrix}] [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 4 \\ - 8 \end{matrix}]$

2x + y = 0, – x + z = 4, – y – 2z = – 8

$\Rightarrow x = - 1, y = 2, z = 3$

$∴ A = [\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 3 \\ - 2 & - 3 & 0 \end{matrix}]$

$∴ 2 (A + I) = [\begin{matrix} 2 & - 2 & 4 \\ 2 & 2 & 6 \\ - 4 & - 6 & 2 \end{matrix}]$

$∴ \det (adj (2 (A + I))) = | 2 (A + I) |^{2} = {(120)}^{2}$

$2^{α} 3^{β} 5^{γ} = 2^{6} \times 3^{2} \times 5^{2}$

$∴ α = 6, β = 2, γ = 2$

$∴ α^{2} + β^{2} + γ^{2} = 36 + 4 + 4 = 44$ .

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