Skip to content
dronacharya.ai
← JEE Advanced 2022 · Paper 2
MathematicsInteger+3 / -1Question 6 of 54

JEE Advanced 2022 · Paper 2, Mathematics question 6

Let β\beta be a real number. Consider the matrix
A=(β0121−231−2).A = \begin{pmatrix}\beta & 0 & 1\\ 2 & 1 & -2\\ 3 & 1 & -2\end{pmatrix}.
If A7−(β−1)A6−βA5A^7 - (\beta-1)A^6 - \beta A^5 is a singular matrix, then the value of 9β9\beta is _____.

Answer and worked solution

Official answer: 3

Factor.
A7−(β−1)A6−βA5=A5(A2−(β−1)A−βI)=A5(A−βI)(A+I).A^7-(\beta-1)A^6-\beta A^5=A^5\left(A^2-(\beta-1)A-\beta I\right)=A^5(A-\beta I)(A+I).
It is singular exactly when one of det⁡A\det A, det⁡(A−βI)\det(A-\beta I), det⁡(A+I)\det(A+I) is zero.

The three determinants.

  • det⁡A=β(−2+2)−0+1(2−3)=−1\det A=\beta(-2+2)-0+1(2-3)=-1.
  • det⁡(A+I)=∣β+10122−231−1∣=(β+1)(−2+2)+1(2−6)=−4\det(A+I)=\begin{vmatrix}\beta+1&0&1\\2&2&-2\\3&1&-1\end{vmatrix}=(\beta+1)(-2+2)+1(2-6)=-4.
  • det⁡(A−βI)=∣00121−β−231−2−β∣=2−3(1−β)=3β−1\det(A-\beta I)=\begin{vmatrix}0&0&1\\2&1-\beta&-2\\3&1&-2-\beta\end{vmatrix}=2-3(1-\beta)=3\beta-1.

Only the last can vanish: β=13\beta=\frac13, so 9β=39\beta=3.

Topics:Determinants and propertiesMatrix algebra
← Previous questionNext question →