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QJEEM25Mab7cd: JEE Main 2025 Mathematics - Class 12 JEEM25Mab7cd Solution: Let A be a matrix such that for all nonzero matrices . If , and …

Let A be a 3×33 \times 3 matrix such that XTAX=O\mathrm{X}^{\mathrm{T}} \mathrm{AX}=\mathrm{O} for all nonzero 3×13 \times 1 matrices X=[xyz]X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] . If A[111]=[14−5],A[121]=[04−8]\mathrm{A}\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}1 \\ 4 \\ -5\end{array}\right], \mathrm{A}\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 4 \\ -8\end{array}\right] , and det⁡(adj⁡(2( A+I)))=2α3β5γ,α,β,γ∈N\operatorname{det}(\operatorname{adj}(2(\mathrm{~A}+\mathrm{I})))=2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma \in N , then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is

Step-by-Step Verified Solution

amp;XTAX=0amp;(xyz)(a1amp;a2amp;a3b1amp;b2amp;b3c1amp;c2amp;c3)(xyz)=0amp;(xyz)(a1x+a2y+a3zb1x+b2y+b3zc1x+c2y+c3z)=0\begin{aligned}& X^T A X=0 \\& (x y z)\left(\begin{array}{lll}a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \\c_1 & c_2 & c_3\end{array}\right)\left(\begin{array}{l}x \\y \\z\end{array}\right)=0 \\& (x y z)\left(\begin{array}{l}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{array}\right)=0\end{aligned}

amp;x(a1x+a2y+a3z)+y(b1x+b2y+b3z)amp;+z(c1x+c2y+c3z)=0amp;a1=0,b2=0c3=0amp;a2+b1=0,a3+c1=0,b3=c2=0\begin{aligned}& x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right) \\& +z\left(c_1 x+c_2 y+c_3 z\right)=0 \\& a_1=0, b_2=0 c_3=0 \\& a_2+b_1=0, a_3+c_1=0, b_3=c_2=0\end{aligned}

amp;A= skew symm matrix amp;A=(0amp;xamp;y−xamp;0amp;z−yamp;−zamp;0);A=(111)=(14−5)amp;⇒A=(0amp;xamp;y−xamp;0amp;z−yamp;−zamp;0)(111)=(14−5)\begin{aligned}& A=\text { skew symm matrix } \\& A=\left(\begin{array}{ccc}0 & x & y \\-x & 0 & z \\-y & -z & 0\end{array}\right) ; \quad A=\left(\begin{array}{l}1 \\1 \\1\end{array}\right)=\left(\begin{array}{l}1 \\4 \\-5\end{array}\right) \\& \Rightarrow A=\left(\begin{array}{ccc}0 & x & y \\-x & 0 & z \\-y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\1 \\1\end{array}\right)=\left(\begin{array}{l}1 \\4 \\-5\end{array}\right)\end{aligned}

amp;x+y=1amp;−x+z=4amp;y+z=5amp;(0amp;xamp;y−xamp;0amp;z−yamp;−zamp;0)(121)=(14−8)\begin{aligned}& x+y=1 \\& -x+z=4 \\& y+z=5 \\& \left(\begin{array}{ccc}0 & x & y \\-x & 0 & z \\-y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\2 \\1\end{array}\right)=\left(\begin{array}{l}1 \\4 \\-8\end{array}\right)\end{aligned}

2x+y=0amp;x=−1−x+z=4amp;y=2−y−2z=−8amp;z=3\begin{array}{ll}2 x+y=0 & x=-1 \\-x+z=4 & y=2 \\-y-2 z=-8 & z=3\end{array}

amp;A=(0amp;−1amp;21amp;0amp;3−2amp;−3amp;0)amp;2( A+I)=(2amp;−2amp;42amp;2amp;6−2amp;−6amp;2)amp;2( A+I)=120⇒det⁡∣adi⁡(2( A+I))∣amp;=1202=26⋅32⋅52amp;α=6,β=2,γ=2\begin{aligned}& A=\left(\begin{array}{ccc}0 & -1 & 2 \\1 & 0 & 3 \\-2 & -3 & 0\end{array}\right) \\& 2(\mathrm{~A}+\mathrm{I})=\left(\begin{array}{ccc}2 & -2 & 4 \\2 & 2 & 6 \\-2 & -6 & 2\end{array}\right) \\& 2(\mathrm{~A}+\mathrm{I})=120 \Rightarrow \operatorname{det}|\operatorname{adi}(2(\mathrm{~A}+\mathrm{I}))| \\& =120^2=2^6 \cdot 3^2 \cdot 5^2 \\& \alpha=6, \beta=2, \gamma=2\end{aligned}

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