Question:
If
\[
A=
\begin{bmatrix}
0&\alpha&\beta\\
\beta&\alpha&0\\
\alpha&0&\beta
\end{bmatrix}
\]
where \(\beta>\alpha>0\) and
\[
AA^T=
\begin{bmatrix}
25&a&b\\
a&25&12\\
b&a&25
\end{bmatrix}
\]
then
\[
a+b+\alpha-\beta=
\]
If
\[
A=
\begin{bmatrix}
0&\alpha&\beta\\
\beta&\alpha&0\\
\alpha&0&\beta
\end{bmatrix}
\]
where \(\beta>\alpha>0\) and
\[
AA^T=
\begin{bmatrix}
25&a&b\\
a&25&12\\
b&a&25
\end{bmatrix}
\]
then
\[
a+b+\alpha-\beta=
\]
Show Hint
For \(AA^T\) problems compare diagonal entries first. They usually give quadratic equations for unknown parameters.
Updated On: Jun 15, 2026
- \(\sqrt{24}\)
- 26
- 25
- 27
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
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 \[ \beta>\alpha \] 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 \[ \boxed{26} \]
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 \[ \beta>\alpha \] 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 \[ \boxed{26} \]
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