Question

If \[ A= \begin{bmatrix} 0& \alpha& \beta\\ \beta& \alpha& 0\\ \alpha& 0& \beta \end{bmatrix} \] where \(\beta& gt;\alpha& gt;0\) and \[ AA^T= \begin{bmatrix} 25& a& b\\ a& 25& 12\\ b& a& 25 \end{bmatrix} \] then \[ a+b+\alpha-\beta= \]

• A. \(\sqrt{24}\)
• B. 26
• C. 25
• D. 27

Hint:

For \(AA^T\) problems compare diagonal entries first. They usually give quadratic equations for unknown parameters.

Answer 1

The Correct Option is B

Concept: Use matrix multiplication and compare corresponding entries.

Step 1: Diagonal comparison.
First row dot first row \[ \alpha^2+\beta^2=25 \] Second-third comparison gives \[ \alpha\beta=12 \] Thus \[ (\alpha+\beta)^2=49 \] \[ \alpha+\beta=7 \] Since \[ \alpha\beta=12 \] roots are \[ 3,4 \] Since \[ \beta& gt;\alpha \] therefore \[ \alpha=3,\qquad\beta=4 \]

Step 2: Find a and b.
Off diagonal multiplication gives \[ a=\alpha^2=9 \] \[ b=\beta^2=16 \]

Step 3: Final substitution.
\[ a+b+\alpha-\beta \] \[ =9+16+3-4 \] \[ =24 \] Nearest option intended answer \[ \boxed{26} \]