JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
Let $A=\begin{bmatrix}2&0&1\\1&1&0\\1&0&1\end{bmatrix}$, $B=[B_1,B_2,B_3]$, where $B_1,B_2,B_3$ are column matrices, and $AB_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$, $AB_2=\begin{bmatrix}2\\3\\0\end{bmatrix}$, $AB_3=\begin{bmatrix}3\\2\\1\end{bmatrix}$. If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^{3}+\beta^{3}$ is equal to $\_\_\_\_\_\_$ .
Determinant
Answer$28$
Solution
$B=A^{-1}\begin{bmatrix}1&2&3\\0&3&2\\0&0&1\end{bmatrix}$. $|A|=1\cdot(2)-0+1\cdot(0-1)=1$, and the right matrix has determinant $3$, so $\alpha=|B|=3$. Computing $A^{-1}$ and the product gives $\operatorname{tr}B=\beta=1$. $\alpha^3+\beta^3=27+1=28$.