Question ID: #846
Let $A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$ and B be a matrix such that $B(I – A) = I + A$. Then the sum of the diagonal elements of $B^T B$ is equal to
Solution:
Given matrix A is skew-symmetric, i.e., $A^T = -A$.
The relation is $B = (I + A)(I – A)^{-1}$.
We need to calculate $B^T B$.
$$ B^T = [(I + A)(I – A)^{-1}]^T $$
$$ B^T = ((I – A)^{-1})^T (I + A)^T $$
$$ B^T = ((I – A)^T)^{-1} (I^T + A^T) $$
$$ B^T = (I – A^T)^{-1} (I + A^T) $$
Since $A^T = -A$:
$$ B^T = (I – (-A))^{-1} (I + (-A)) $$
$$ B^T = (I + A)^{-1} (I – A) $$
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):
$(I – A)(I + A) = I – A^2 = (I + A)(I – A)$.
We can rearrange the terms:
$$ B^T B = (I + A)^{-1} (I + A) (I – A) (I – A)^{-1} $$
$$ B^T B = I \cdot I = I $$
$B^T B$ is the identity matrix of order 3.
The sum of diagonal elements (Trace) = $1 + 1 + 1 = 3$.
Ans. (3)
Was this solution helpful?
YesNo