JEE Main 2026 Jan 23 Shift 2, Mathematics Q24: Algebra Of Matrices

JEE Main2026Jan 23, Shift 2Mathematics

Let $A = 02 - 3 - 201 3 - 10$ and $B$ be a matrix such that $B (I - A) = I + A$ . Then the sum of the diagonal elements of $B^{T} B$ is equal to _______.

Your answer

Solution

$A$ is skew-symmetric: $A^{T} = - A$ . From $B (I - A) = I + A$ :

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

Then $B^{T} = [(I - A)^{- 1}]^{T} (I + A)^{T} = (I - A^{T})^{- 1} (I + A^{T}) = (I + A)^{- 1} (I - A)$ .

Now compute $B^{T} B$ :

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

Since $(I - A)$ and $(I + A)$ commute (both are polynomials in $A$ ), $(I - A) (I + A) = (I + A) (I - A)$ , so

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

Sum of diagonal entries of $I_{3 \times 3}$ is $3$ .

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JEE Main 2026 Jan 23 Shift 2, Mathematics Q24: Algebra Of Matrices

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