MCQ
If $A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], A ^{-1}=\alpha A+\beta I$ and $\alpha+\beta=-2$ then $4 \alpha^2+\beta^2+\lambda^2$ is equal to:
- A$12$
- B$10$
- C$19$
- ✓$14$
$\Rightarrow x^2-11 x+10-5 \lambda=0$
$\Rightarrow(10-5 \lambda) A^{-1}=-A+11 I$
$\alpha+\beta=-2 \Rightarrow \frac{10}{10-5 \lambda}=-2 \Rightarrow 10-5 \lambda=-5 \Rightarrow \lambda=3$
$\therefore \alpha=\frac{1}{5} \quad \quad \beta=\frac{-11}{5}$
$\therefore 4 a^2+\beta^2+\lambda^2=\frac{4}{25}+\frac{121}{25}+3^2=14 \text { Ans. }$
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