JEE Advanced 2022 Paper 2 · Q06 · Determinants

Question

Let $\beta$ be a real number. Consider the matrix $A=\begin{pmatrix}\beta & 0 & 1\ 2 & 1 & -2\ 3 & 1 & -2\end{pmatrix}$. If $A^{7}-(\beta-1)A^{6}-\beta A^{5}$ is a singular matrix, then the value of $9\beta$ is _____.

SolveKar AI · Accepted Answer

$A^7-(\beta-1)A^6-\beta A^5 = A^5[A^2-(\beta-1)A-\beta I] = A^5(A-\beta I)(A+I)$. The matrix is singular iff $\det A^5 \det(A-\beta I)\det(A+I)=0$. Compute $\det A$ by expanding: $\beta(1\cdot(-2)-(-2)\cdot 1) - 0 + 1\cdot(2-3) = \beta(0)+(-1) = -1$, so $\det A^5
e 0$. Compute $\det(A+I)$: $\begin{vmatrix}\beta+1&0&1\ 2&2&-2\ 3&1&-1\end{vmatrix}=(\beta+1)(2\cdot(-1)-(-2)\cdot 1)-0+1(2-6)=(\beta+1)(0)-4=-4
e 0$. Hence $\det(A-\beta I)=0$: $\begin{vmatrix}0&0&1\ 2&1-\beta&-2\ 3&1&-2-\beta\end{vmatrix}=1\cdot(2-3(1-\beta))=1\cdot(3\beta-1)=0$, giving $\beta=1/3$. Therefore $9\beta=3$.

JEE Advanced 2022 Paper 2 · Q06 · Determinants

Solution