Question

If \( a, b, c \in \mathbb{R} \), \( a < b < c \), \( A = \begin{bmatrix} -c & 0 & c 0 & -b & b a & a & 0 \end{bmatrix} \), \( 12A^{-1} = \begin{bmatrix} -b & c & bc b & -c & bc b & c & bc \end{bmatrix} \) and Trace of \( A = -5 \), then \( a^2 + b^2 + c^2 = \)

• A. 29
• B. 21
• C. 26
• D. 14

Hint:

To avoid calculating the full adjoint, compare just one or two corresponding elements. For example, the element at (3,3) in the given \( 12A^{-1} \) is \( bc \). In the computed \( 12A^{-1} \), it is \( 12 \times (1/2a) = 6/a \). Setting \( 6/a = bc \) quickly connects the variables.

Answer 1

The Correct Option is D