NVAMediumJEE 2026Algebra of Matrices

JEE Mathematics 2026 Question with Solution

Let A=[023201310]A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} and BB be a matrix such that B(IA)=I+AB(I - A) = I + A. Then the sum of the diagonal elements of BTBB^T B is equal to _____

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given:

A=[023201310]A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}

and B(IA)=I+AB(I-A)=I+A.

Find: The sum of the diagonal elements of BTBB^T B.

From

B(IA)=I+AB(I-A)=I+A

we get

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

Notice that AA is a real skew-symmetric matrix, so

AT=AA^T=-A

Now compute BTB^T:

BT=[(IA)1]T(I+A)T=(IAT)1(I+AT)\begin{aligned} B^T &= \left[(I-A)^{-1}\right]^T (I+A)^T \\ &= (I-A^T)^{-1}(I+A^T) \end{aligned}

Using AT=AA^T=-A,

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

Also,

(I+A)(IA)=IA2=(IA)(I+A)(I+A)(I-A)=I-A^2=(I-A)(I+A)

so I+AI+A and IAI-A commute.

Therefore,

BTB=(I+A)1(IA)(I+A)(IA)1=(I+A)1(I+A)(IA)(IA)1=I\begin{aligned} B^T B &= (I+A)^{-1}(I-A)(I+A)(I-A)^{-1} \\ &= (I+A)^{-1}(I+A)(I-A)(I-A)^{-1} \\ &= I \end{aligned}

Thus BTBB^T B is the identity matrix of order 33. Hence the sum of its diagonal elements is

1+1+1=31+1+1=3

Therefore, the required answer is 33.

Cayley Transform Observation

Given: AA is real skew-symmetric and B=(I+A)(IA)1B=(I+A)(I-A)^{-1}.

Find: The sum of the diagonal elements of BTBB^T B.

For a real skew-symmetric matrix, the Cayley transform

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

is an orthogonal matrix. Hence

BTB=IB^T B=I

Since the identity matrix is of order 33, its diagonal sum is

tr(I)=3\operatorname{tr}(I)=3

Therefore, the required answer is 33.

Common mistakes

  • Assuming one must multiply out BB explicitly. That is unnecessary and can make the work lengthy. Instead, use the relation B=(I+A)(IA)1B=(I+A)(I-A)^{-1} together with the skew-symmetry of AA.

  • Forgetting that AT=AA^T=-A for a skew-symmetric matrix. If this is missed, the transpose of BB is computed incorrectly. First identify the matrix type, then use AT=AA^T=-A carefully.

  • Concluding that BTB=IB^T B=I without justifying why the factors can be rearranged. The rearrangement works because (I+A)(IA)=IA2=(IA)(I+A)(I+A)(I-A)=I-A^2=(I-A)(I+A), so the two matrices commute.

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