Let \(\mathrm{A}=\left[\begin{array}{ccc}0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0\end{array}\right]\) and B be a matrix such that \(\mathrm{B}(\mathrm{I}-\mathrm{A})=\mathrm{I}+\mathrm{A}\). Then the sum of the diagonal elements of \(B^{T} B\) is equal to

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Matrices

Let

A = 0 - 2 3 20 - 1 - 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

Answer: 3

The matrix

A

satisfies

A^{T} = - A

, so it is skew-symmetric. Solving the relation,

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

, the Cayley transform of a skew-symmetric matrix. Its transpose is

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

. Since

I - A

and

I + A

are polynomials in

A

, they commute, so

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

. The sum of the diagonal elements of

B^{T} B

is the trace of the

3 \times 3

identity matrix, which is

3

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Integer answer