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Mathematics · Matrices

JEE Main 2025 — 24 January, Morning Shift — Question 24

Let be a 3×33 \times 3 matrix such that XTAX=OX^T A X = O for all nonzero 3×13 \times 1 matrices X=[xyz].X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} .

If A[111]=[14−5],A[121]=[04−8], and\text{If } A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}, A \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}, \text{ and} det (adj(2(A+I)))=2α3β5γ,α,β,γ∈N, then\text{det } (\text{adj} (2(A + I))) = 2^\alpha 3^\beta 5^\gamma , \alpha, \beta, \gamma \in N, \text{ then} α2+β2+γ2 is\alpha^2 + \beta^2 + \gamma^2 \text{ is}

Answer: 44

Numerical answer — enter this value.

Step-by-step solution

XTAX=0X^T A X = 0 (xyz)(a1a2a3b1b2b3c1c2c3)(xyz)=0(xyz) \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 (xyz)(a1x+a2y+a3zb1x+b2y+b3zc1x+c2y+c3z)=0(xyz) \begin{pmatrix} a_1 x + a_2 y + a_3 z \\ b_1 x + b_2 y + b_3 z \\ c_1 x + c_2 y + c_3 z \end{pmatrix} = 0 x(a1x+a2y+a3z)+y(b1x+b2y+b3z)+z(c1x+c2y+c3z)=0x(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 a1=0,b2=0,c3=0a_1 = 0, b_2 = 0, c_3 = 0 a2+b1=0,a3+c1=0,b3+c2=0a_2 + b_1 = 0, a_3 + c_1 = 0, b_3 + c_2 = 0 A=skew symm matrixA = \text{skew symm matrix} A=(0xy−x0z−y−z0);A(111)=(14−5)A = \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} ; A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -5 \end{pmatrix} ⇒A=(0xy−x0z−y−z0)(111)=(14−5)\Rightarrow A = \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -5 \end{pmatrix} x+y=1x + y = 1 −x+z=4-x + z = 4 −y−z=5-y - z = 5 A(121)=(0xy−x0z−y−z0)(121)=(14−8)A \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -8 \end{pmatrix} 2x+y=0x=−12x + y = 0 \quad x = -1 −x+z=4y=2-x + z = 4 \quad y = 2 −y−2z=−8z=3-y - 2z = -8 \quad z = 3 A=(0−12103−2−30)A = \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix} 2(A+I)=(2−24226−4−62)2(A + I) = \begin{pmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{pmatrix} 2(A+I)=120⇒det⁡∣adj(2(A+I))∣=1202=26⋅32⋅522(A + I) = 120 \Rightarrow \det | \text{adj} (2(A + I)) | = 120^2 = 2^6 \cdot 3^2 \cdot 5^2 α=6,β=2,γ=2\alpha = 6, \beta = 2, \gamma = 2 α2+β2+γ2=44\alpha^2 + \beta^2 + \gamma^2=44

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Mathematics
Chapter
Matrices
Topic
Adjoint of a Square Matrix

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