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3 and 4 .Determinants and Matrices
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Let be a $3 \times 3$ matrix such that $X ^{ T } AX = O$ for all nonzero $3 \times 1$ matrices $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$. If $A \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}1 \\ 4 \\ -5\end{array}\right], A \left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\ 4 \\ -8\end{array}\right]$, and $\operatorname{det}(\operatorname{adj}(2(A+ I )))=2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma, \in N$, then $\alpha^2+\beta^2+\gamma^2$ is
A$42$
B$43$
C$45$
D$44$
(JEE MAIN-2025)
Solution
$X ^{ T } AX =0$
(xyz) $\left(\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=0$
(xyz) $\left(\begin{array}{l}a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z\end{array}\right)=0$
$x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right)$
$+z\left(c_1 x+c_2 y+c_3 z\right)=0$
$a_1=0, b_2=0 c_3=0$
$a_2+b_1=0, a_3+c_1=0, b_3=c_2=0$
$A=\text { skew symm matrix }$
$A=\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right) ; \quad A=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -5\end{array}\right)$
$\Rightarrow A=\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -5\end{array}\right)$
$x+y=1$
$-x+z=4$
$y+z=5$
$\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\ 2 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -8\end{array}\right)$
$2 x+y=0 x=-1$
$-x+z=4 y=2$
$-y-2 z=-8 z=3$
$A=\left(\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right)$
$2(A+ I )=\left(\begin{array}{ccc}2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2\end{array}\right)$
$\begin{array}{l}2(A+ I )=120 \Rightarrow \operatorname{det}|\operatorname{adi}(2(A+ I ))| \\ =120^2=2^6 \cdot 3^2 \cdot 5^2 \\ \alpha=6, \beta=2, \gamma=2\end{array}$
(xyz) $\left(\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=0$
(xyz) $\left(\begin{array}{l}a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z\end{array}\right)=0$
$x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right)$
$+z\left(c_1 x+c_2 y+c_3 z\right)=0$
$a_1=0, b_2=0 c_3=0$
$a_2+b_1=0, a_3+c_1=0, b_3=c_2=0$
$A=\text { skew symm matrix }$
$A=\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right) ; \quad A=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -5\end{array}\right)$
$\Rightarrow A=\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -5\end{array}\right)$
$x+y=1$
$-x+z=4$
$y+z=5$
$\left(\begin{array}{ccc}0 & x & y \\ -x & 0 & z \\ -y & -z & 0\end{array}\right)\left(\begin{array}{l}1 \\ 2 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 4 \\ -8\end{array}\right)$
$2 x+y=0 x=-1$
$-x+z=4 y=2$
$-y-2 z=-8 z=3$
$A=\left(\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right)$
$2(A+ I )=\left(\begin{array}{ccc}2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2\end{array}\right)$
$\begin{array}{l}2(A+ I )=120 \Rightarrow \operatorname{det}|\operatorname{adi}(2(A+ I ))| \\ =120^2=2^6 \cdot 3^2 \cdot 5^2 \\ \alpha=6, \beta=2, \gamma=2\end{array}$
Standard 12
Mathematics
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