Jee Main 2024 27 01 2024 Shift 1 - Question20

Answer: (117)

Solution:

$B=\left[\begin{array}{lll}x _1 & x _2 & x _3 \\ y _1 & y _2 & y _3 \\ z _1 & z _2 & z _3\end{array}\right], A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$ $\Rightarrow B _1=\left[\begin{array}{l}x _1 \\ y _1 \\ z _1\end{array}\right], B _2=\left[\begin{array}{l}x _2 \\ y _2 \\ z _2\end{array}\right], B _3=\left[\begin{array}{l}x _3 \\ y _3 \\ z _3\end{array}\right]$

$A \cdot B _1=\left[\begin{array}{c}2 x _1+y _1 \\ z _1 \\ x _1\end{array}\right]=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right] \Rightarrow 2 x _1+y _1=2$

$z _1=3$

$x _1=1$

$\Rightarrow x _1=1, y _1=0, z _1=3$

$A \cdot B _2=\left[\begin{array}{c}2 x _2+y _2 \\ z _2 \\ x _2\end{array}\right]=\left[\begin{array}{l}2 \\ 0 \\ 0\end{array}\right]$

$\Rightarrow 2 x _2+y _2=2, z _2=0, x _2=0$

$x _2=0$

$Z _2=0$

$y _2=2$

$A \cdot B _3=\left[\begin{array}{c}2 x _3+y _3 \\ z _3 \\ x _3\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$

$2 x _3+y _3=3, z _3=2, x _3=1$

$x _3=1$

$z _3=2$

$y _3=1$

$\Rightarrow B=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 1 \\ 3 & 0 & 2\end{array}\right]$

$\alpha=|B|=4-6=-2$

$\beta=5$

$\alpha^{3}+\beta^{3}=-8+125$

$=117$