Question

Let $A$ be a matrix of order $2\times 2$ such that $A^2=0$ then $A^2-(a+d)A+(ad-bc)I$ is equal to

• A. $I$
• B. $0_{2\times 2}$
• C. $-I$
• D. none of these

Answer 1

The correct option is B $0_{2\times 2}$

Given $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

$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& ab+bd\\ ca+cd& bc+d^2\end{array}\right)............................\left(1\right)$

$-(a+d)A=\left(\begin{array}{cc}-a^2-ad& -ab-bd\\ -ac-cd& -ad-d^2\end{array}\right)..............\left(2\right)$

$(ad-bc)I=\left(\begin{array}{cc}ad-bc& 0\\ 0& ad-bc\end{array}\right).............\left(3\right)$

Adding 1,2,3 we get,

$A^2-(a+d)A+(ad-bc)I=\left(\begin{array}{cc}{a}^2+bc-a^2-ad+ad-bc& ab+bd-ab-bd\\ ac+cd-ac-cd& bc+d^2-ad-d^2+ad-bc\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$

Hence, the value of $A^2-(a+d)A+(ad-bc)I=0_{2\times 2}$