MCQ
Let $A=\left(\begin{array}{cc}4 & -2 \\ \alpha & \beta\end{array}\right)$ . If $A ^{2}+\gamma A +18 I = O$, then $\operatorname{det}( A )$ is equal to
  • A
    $-18$
  • $18$
  • C
    $-50$
  • D
    $50$

Answer

Correct option: B.
$18$
b
The characteristic equation for $A$ is $| A -\lambda I |=0$

$\left|\begin{array}{cc}4-\lambda & -2 \\ \alpha & \beta-\lambda\end{array}\right|=0$

$(4-\lambda)(\beta-\lambda)+2 \alpha=0$

$\lambda^{2}-(\beta+4) \lambda+4 \beta+2 \alpha=0$

Put $\lambda= A$

$A^{2}-(\beta+4) A+(4 \beta+2 \alpha) I=0$

On comparison $-9(\beta+4)=\gamma \& 4 \beta+2 \alpha=18$

and $| A |=4 \beta+2 \alpha=18$

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