Matrices & DeterminantsmediumFree

Matrices & Determinants — JEE Maths practice question

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

Question

A

and

B

are two non-singular matrices of order 3 such that

AA^T = 2I

and

A^{-1} = A^T - A\,\text{adj}(2B^{-1})

, then

\det(B)

equals:

4

4\sqrt{2}

16

16\sqrt{2}

correct

Solution

Step 1: Take the determinant of

AA^T = 2I

|A|^2 = |2I| = 2^3 = 8

Step 2: Multiply the given relation on the left by

A

A A^{-1} = AA^T - A^2\,\text{adj}(2B^{-1}) \implies I = 2I - A^2\,\text{adj}(2B^{-1})

A^2\,\text{adj}(2B^{-1}) = I

Step 3: Take the determinant

|A|^2\,|\text{adj}(2B^{-1})| = 1 \implies 8\,|\text{adj}(2B^{-1})| = 1

Using

|\text{adj}(M)| = |M|^2

for order 3 and

|2B^{-1}| = \dfrac{2^3}{|B|}

8\cdot\left(\frac{8}{|B|}\right)^2 = 1 \implies \frac{512}{|B|^2} = 1 \implies |B|^2 = 512 \implies |B| = 16\sqrt{2}

Answer: (4)

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