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Matrices & Determinants: Let Non Singular Matrices Order 6ab Pmatrix Pmatrix

JEE Maths question with a full step-by-step solution.

Question
Let AA and BB be two non-singular matrices of order 2 such that
9A2B−6AB+B=9{B−(2−321)}9A^2B - 6AB + B = 9\left\{B - \begin{pmatrix} 2 & -3 \\ 2 & 1 \end{pmatrix}\right\}
 If B=(3011)B = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} and Tr(A)=103\text{Tr}(A) = \dfrac{10}{3}, then the value of det⁡(A)\det(A) can be:
A11
B22correct
C33
D44
Solution
Step 1: Rearrange the given relation
9A2B−6AB+B=9B−9(2−321)  ⟹  (9A2−6A)B=8B−9(2−321)9A^2B - 6AB + B = 9B - 9\begin{pmatrix} 2 & -3 \\ 2 & 1 \end{pmatrix} \implies (9A^2 - 6A)B = 8B - 9\begin{pmatrix} 2 & -3 \\ 2 & 1 \end{pmatrix}
 Step 2: Evaluate the right-hand side
8B=(24088),9(2−321)=(18−27189)8B = \begin{pmatrix} 24 & 0 \\ 8 & 8 \end{pmatrix}, \qquad 9\begin{pmatrix} 2 & -3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 18 & -27 \\ 18 & 9 \end{pmatrix}
(9A2−6A)B=(627−10−1)(9A^2 - 6A)B = \begin{pmatrix} 6 & 27 \\ -10 & -1 \end{pmatrix}
 Step 3: Multiply on the right by B−1B^{-1} With B−1=13(10−13)B^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix}:
9A2−6A=(627−10−1)⋅13(10−13)=(−727−3−1)9A^2 - 6A = \begin{pmatrix} 6 & 27 \\ -10 & -1 \end{pmatrix}\cdot\frac{1}{3}\begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} -7 & 27 \\ -3 & -1 \end{pmatrix}
 Step 4: Form a perfect square
9A2−6A+I=(3A−I)2=(−627−30)9A^2 - 6A + I = (3A - I)^2 = \begin{pmatrix} -6 & 27 \\ -3 & 0 \end{pmatrix}
∣3A−I∣2=det⁡(−627−30)=81  ⟹  ∣3A−I∣=±9|3A - I|^2 = \det\begin{pmatrix} -6 & 27 \\ -3 & 0 \end{pmatrix} = 81 \implies |3A - I| = \pm 9
 Step 5: Express the determinant in terms of det⁡A\det A and Tr(A)\text{Tr}(A)
∣3A−I∣=9det⁡A−3 Tr(A)+1=9det⁡A−10+1=9det⁡A−9|3A - I| = 9\det A - 3\,\text{Tr}(A) + 1 = 9\det A - 10 + 1 = 9\det A - 9
9det⁡A−9=±9  ⟹  det⁡A=2 or 09\det A - 9 = \pm 9 \implies \det A = 2 \text{ or } 0
 Since AA is non-singular, det⁡A=2\det A = 2. Answer: (2)
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