JEE Advanced 2022 Paper 2, Mathematics Section 1 Q6: Properties Of Determinant

Factor: $A^7-(\beta -1)A^6-\beta A^5=A^5\left[A^2-(\beta -1)A-\beta I\right]=A^5(A+I)(A-\beta I)$.

Compute $\det (A)$ by expansion along the first row: $\det (A)=\beta (1\cdot (-2)-(-2)\cdot 1)-0+1\cdot (2-3)=\beta (0)+(-1)=-1$.

So $\det (A^5)=(-1{)}^5=-1\ne 0$. The product is singular when $\det (A+I)=0$ or $\det (A-\beta I)=0$.

$\det (A+I)=\left|\begin{array}{c}\beta +101\\ 22-2\\ 31-1\end{array}\right|=(\beta +1)(-2+2)-0+1(2-6)=-4\ne 0$.

$\det (A-\beta I)=\left|\begin{array}{c}001\\ 21-\beta -2\\ 31-2-\beta \end{array}\right|$. Expanding along the first row: $1\cdot (2\cdot 1-3(1-\beta ))=2-3+3\beta =3\beta -1$.

Setting $3\beta -1=0$ gives $\beta =\frac{1}{3}$, so $9\beta =3$.