First slide

Applications of determinant

Question

$If A and B are two non-singular matrices of order 3 such that {AA}^{T} = 2 I and$ $A^{- 1} = A^{T} - A . adj (2 B^{- 1}), then \det (B) equals$

Moderate

A

$\pm 8$
B

$\pm 8 \sqrt{2}$
C

$\pm 16 \sqrt{2}$
D

$\pm 32$

Solution

$\begin{array}{l} Given that {AA}^{T} = 2 I \\ \Rightarrow | A |^{2} = 8 \\ And A^{- 1} = A^{T} - A adj (2 B^{- 1}) \\ \Rightarrow I = {AA}^{T} - A^{2} adj (2 B^{- 1}) \\ \Rightarrow I = 2 I - A^{2} adj (2 B^{- 1}) \\ \Rightarrow A^{2} adj (2 B^{- 1}) = I \\ \Rightarrow |A^{2} ∥ adj (2 B^{- 1})| = 1 \\ \Rightarrow 8 {|2 B^{- 1}|}^{2} = 1 ∵ |a d j A| = {|A|}^{n - 1} \\ \Rightarrow 8 \times \frac{64}{| B |^{2}} = 1 ∵ |B^{- 1}| = \frac{1}{|B|} a n d |k A| = k^{3} |A| = {(2^{3} |B^{- 1}|)}^{2} \\ \Rightarrow | B |^{2} = 64 \times 8 \\ ∴ | B | = \pm 16 \sqrt{2} \end{array}$

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