Question

If A = $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ (where $b\ne c$) and satisfies the equation $A^2+kI=0$, then

• A. $a+d=0$
• B. $k=-\left|A\right|$
• C. $k=\left|A\right|$
• D. None of the above

Answer 1

Answer: (C)

$k=\left|A\right|$

We have,

$A^2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}{a}^2+bc& ab+db\\ ac+cd& bc+d^2\end{array}\right]$

As A satisfies $x^2+k=0$ therefore

$A^2+kI=0$

$\Rightarrow \left[\begin{array}{cc}{a}^2+bc+k& (a+d)b\\ (a+d)c& bc+d^2+k\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow a^2+bc+k=0,bc+d^2+k=0$

and $(a+d)b=(a+d)c=0$

As $bc\ne 0,b\ne 0,c\ne 0$ , so

$a+d=0$

$\Rightarrow a=-d$

Also,

$k=-(a^2+bc)$

$=-(d^2+bc)$

$=-\left(\right(-ad)+bc)$

$=|A|$