Question

$IfA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right](wherebc\ne 0)satisfiestheequation{x}^2+k=0,then$

• A. $a-d=0$
• B. $k=-|A|$
• C. $k=|A|$
• D. $ad=bc$

Answer 1

The correct option is C

$k=|A|$

$Asatisfies{x}^2+k=0,hence,A^2+kI=0A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],A^2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right]A^2+kI=0\Rightarrow \left[\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right]+\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]b(a+d)=0,c(a+d)=0Alsogiventhatbc\ne 0.Hencea+d=0ord=-aSo,a^2+bc=-ad+bc=d^2+bc$

Equating the first and fourth elements of the LHS to zero,
$-ad+bc+k=0\Rightarrow k=ad-bc=|A|$