Aspire Faculty ID #18034 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

Let \( A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} \) and \( B \) be a matrix such that \( B(I - A) = I + A \)

Solution

$A^T = -A$

$B = (I + A)(I - A)^{-1}$

$B^T = \left((I - A)^{-1}\right)^T (I + A)^T$

$= (I - A^T)^{-1}(I + A)^T$

$= (I + A)^{-1}(I - A)$

Now

$B^T B = (I + A)^{-1}(I - A)(I + A)(I - A)^{-1}$

$= (I + A)^{-1}(I + A)(I - A)(I - A)^{-1}$

$= I$

$\Rightarrow \text{tr}(B^T B) = 3$

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