JEE Advanced 2021 Paper 1 · Q07 · Determinants

Correct answer:1

Solution

The system is consistent iff (eq1)+(eq3)-(eq2)=0, i.e., $\alpha-2\beta+\gamma-1=0$, so plane $P:\ x-2y+z-1=0$ in $(\alpha,\beta,\gamma)$ space. The matrix $M=\begin{pmatrix}\alpha&2&\gamma\\\beta&1&0\\-1&0&1\end{pmatrix}$ has $|M|=\alpha(1)-2(\beta-0)+\gamma(0+1)=\alpha-2\beta+\gamma$. From consistency condition $\alpha-2\beta+\gamma=1$, so $|M|=1$.

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