Question

If $ad-bc\ne 0$ and $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $A^2+xA+y{I}_2=0$ then

• A. $x=-(a+b)$
• B. $x=-(a+d)$
• C. $y=ad-bc$
• D. $y=bc-ad$

Answer 1

Answer: (B), (C)

$x=-(a+d)$
$y=ad-bc$

$A^2=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]$
but $A^2+xA+y{I}_2=O$ that implies
$a^2+bc+xa+y=0$-----(1)
$ab+bd+xb=0$-----(2)
$ac+cd+xc=0$-----(3)
$bc+d^2+xd+y=0$-----(4)
As$ad-bc\ne 0$ one of $b,d$ is not zero.
Assuming, $b\ne 0$.
From (2) we get $x=-(a+d)$. Substituting this in (1), we obtain $y=ad-bc$
Hence, option B and C.