Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x\end{array}\right|=0$.

Home » Show that is a root of the equation .

Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x\end{array}\right|=0$ .

Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x\end{array}\right|=0$ .

Solution:

Operating $R_{1} \rightarrow R_{1}-R_{2}$
$\begin{aligned} 0 &=\left|\begin{array}{ccc} x-2 & -6+3 x & -1-x+3 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x \end{array}\right| \\ 0 &=\left|\begin{array}{ccc} x-2 & 3(x-2) & -(x-2) \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x \end{array}\right| \end{aligned}$
Taking $(x-2)$ common from $R_{1}$
$0=(x-2)\left|\begin{array}{ccc} 1 & 1 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x \end{array}\right|$
Here, we can see that $x-2$ is a factor of the determinant.
We can say that when $x-2$ is put in the equation, we get 0 .
$x-2=0$
$\rightarrow \mathrm{x}=2$