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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
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
    $19$
  • ✓
    $14$
✓

Answer

Correct option: D.
$14$
d
$|A-x I|=0 \Rightarrow\left|\begin{array}{cc}1-x & 5 \\ \lambda & 10-x\end{array}\right|=0$

$\Rightarrow x^2-11 x+10-5 \lambda=0$

$\Rightarrow(10-5 \lambda) A^{-1}=-A+11 I$

$\alpha+\beta=-2 \Rightarrow \frac{10}{10-5 \lambda}=-2 \Rightarrow 10-5 \lambda=-5 \Rightarrow \lambda=3$

$\therefore \alpha=\frac{1}{5} \quad \quad \beta=\frac{-11}{5}$

$\therefore 4 a^2+\beta^2+\lambda^2=\frac{4}{25}+\frac{121}{25}+3^2=14 \text { Ans. }$

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