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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

A

and

B

be two non-singular matrices of order 2 such that

9A^2B - 6AB + B = 9\left\{B - \begin{pmatrix} 2 & -3 \\ 2 & 1 \end{pmatrix}\right\}

B = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}

and

\text{Tr}(A) = \dfrac{10}{3}

, then the value of

\det(A)

can be:

1

2

correct

3

4

Solution

Step 1: Rearrange the given relation

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 = \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}

(9A^2 - 6A)B = \begin{pmatrix} 6 & 27 \\ -10 & -1 \end{pmatrix}

Step 3: Multiply on the right by

B^{-1}

With

B^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 0 \\ -1 & 3 \end{pmatrix}

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

9A^2 - 6A + I = (3A - I)^2 = \begin{pmatrix} -6 & 27 \\ -3 & 0 \end{pmatrix}

|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

and

\text{Tr}(A)

|3A - I| = 9\det A - 3\,\text{Tr}(A) + 1 = 9\det A - 10 + 1 = 9\det A - 9

9\det A - 9 = \pm 9 \implies \det A = 2 \text{ or } 0

Since

A

is non-singular,

\det A = 2

. Answer: (2)

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