Let be a $3 \times 3$ matrix such that $\mathrm{X}^{\mathrm{T}} \mathrm{AX}=\mathrm{O}$ for all nonzero $3 \times 1$ matrices $\mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]$. If $\mathrm{A}\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}1 \\ 4 \\ -5\end{array}\right], \mathrm{A}\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\ 4 \\ -8\end{array}\right]$, and $\operatorname{det}(\operatorname{adj}(2(\mathrm{~A}+\mathrm{I})))=2^{\alpha} 3^{\beta} 5^{\gamma}, \alpha, \beta, \gamma, \in \mathrm{N}$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is

Matrices — JEE Main · 2025 · 24 Jan 2025 · Shift 1 Question with Solution

Let be a

3 \times 3

matrix such that

X^{T} AX = O

for all nonzero

3 \times 1

matrices

X = x y z

.

If

A 111 = 14 - 5, A 121 = 04 - 8

, and

det (adj (2 (A + I))) = 2^{α} 3^{β} 5^{γ}, α, β, γ, \in N

, then

α^{2} + β^{2} + γ^{2}

Answer: 44

Since

X^{T} A X = 0

for every nonzero column

X

, the symmetric part of

A

must vanish (only the skew-symmetric part of a matrix contributes zero to every such quadratic form), so

A

is skew-symmetric:

A = 0 - x - y x 0 - z y z 0 .

From

A 111 = 14 - 5

x + y = 1

- x + z = 4

- y - z = - 5 \Rightarrow y + z = 5

.

From

A 121 = 04 - 8

2 x + y = 0

- x + z = 4

(consistent),

- y - 2 z = - 8 \Rightarrow y + 2 z = 8

.

From

y + z = 5

and

y + 2 z = 8

z = 3, y = 2

. From

x + y = 1

x = - 1

(consistent with

2 x + y = - 2 + 2 = 0

and

- x + z = 1 + 3 = 4

A = 01 - 2 - 1 0 - 3 230, A + I = 11 - 2 - 1 1 - 3 231 .

det (A + I) = 1 (1 + 9) + 1 (1 + 6) + 2 (- 3 + 2) = 10 + 7 - 2 = 15.

For a

3 \times 3

matrix,

det (2 M) = 2^{3} det M

, so

det (2 (A + I)) = 8 \times 15 = 120

, and

det (adj (N)) = (det N)^{2}

for

3 \times 3

N

det (adj (2 (A + I))) = 12 0^{2} = 14400 = 2^{6} \cdot 3^{2} \cdot 5^{2} .

α = 6, β = 2, γ = 2

, giving

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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Integer answer