Question

$A$ is a $2\times 2$ matrix such that $A\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]$ and $A^2\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right]$. The sum of the elements of $A$ is

• A. $-1$
• B. $0$
• C. $2$
• D. none of these

Answer 1

The correct option is D none of these

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

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

Now given $A\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]\Rightarrow \left|\begin{array}{cc}a& b\\ c& d\end{array}\right|\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}a-b\\ c-d\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]$

$\Rightarrow a-b=-1.....\left(i\right)$ and $c-d=2.....\left(ii\right)$

Also given $A^2\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right]\Rightarrow \left|\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+dc& bc+d^2\end{array}\right|\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right]$

$\Rightarrow a^2+bc-ab-bd=-1....\left(iii\right)$ and $ac+dc-bc-d^2=0.....\left(iv\right)$

Solving all the four equation we get $a=-3,b=-2,c=4,d=2$

Adding the elements of $A$, we get

$a+b+c+d=1$