Question

$\left|\begin{array}{cc}{x}^2+x& x\\ 3& 2\end{array}\right|+\left|\begin{array}{cc}0& -1\\ -x+1& x\end{array}\right|=\left|\begin{array}{cc}0& -2\\ 5& 1\end{array}\right|$ then x is equal to-

• A. -1
• B. 2
• C. 1
• D. No value of x

Answer 1

The correct option is A -1

Given $\left[\begin{array}{cc}{x}^2+x& x\\ 3& 2\end{array}\right]+\left[\begin{array}{cc}0& -1\\ -x+1& x\end{array}\right]=\left[\begin{array}{cc}0& -2\\ 5& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{x}^2+x& x-1\\ 3-x+1& 2+x\end{array}\right]=\left[\begin{array}{cc}0& -2\\ 5& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{x}^2+x& x-1\\ 4-x& 2+x\end{array}\right]=\left[\begin{array}{cc}0& -2\\ 5& 1\end{array}\right]$

$\Rightarrow x^2+x=0$ …………$\left(1\right)$

$x-1=-2$

$\Rightarrow x=-2+1=-1$

$4-x=5$

$\Rightarrow x=4-5=-1$

$2+x=1$

$\Rightarrow x=1-2=-1$

And substituting $x=-1$ in $\left(1\right)$, $(-1{)}^2+(-1)$

$\Rightarrow 1-1=0$

$\therefore x=-1$.