Question:
If \( A \) is an \( n \times n \) non-singular matrix such that \( AA^T = A^{-1}A \) and \( B = A^{-1}A^T \), then \( BB' \) is equal to
If \( A \) is an \( n \times n \) non-singular matrix such that \( AA^T = A^{-1}A \) and \( B = A^{-1}A^T \), then \( BB' \) is equal to
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Transpose reverses order — very important in matrix simplifications!
Updated On: Apr 22, 2026
- \( I + B \)
- \( I \)
- \( B^{-1} \)
- \( (B^{-1})' \)
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The Correct Option is B
Solution and Explanation
Concept:
Use properties:
\[
(A^T)^T = A,\quad (AB)^T = B^TA^T
\]
Step 1: Given.
\[ B = A^{-1}A^T \]
Step 2: Transpose.
\[ B' = (A^{-1}A^T)^T = A(A^{-1})^T \]
Step 3: Multiply.
\[ BB' = A^{-1}A^T \cdot A(A^{-1})^T = I \]
Step 1: Given.
\[ B = A^{-1}A^T \]
Step 2: Transpose.
\[ B' = (A^{-1}A^T)^T = A(A^{-1})^T \]
Step 3: Multiply.
\[ BB' = A^{-1}A^T \cdot A(A^{-1})^T = I \]
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