Question

Use properties of determinants to solve for x: $\left|\begin{array}{ccc}x+a& b& x\\ c& x+b& a\\ a& b& x+c\end{array}\right|=0$ and $x\ne 0$.

Answer 1

We have

$\Delta =\left|\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right|=0$

Using determinants properties

$C_1\to C_1+C_2+C_3$

$\Delta =\left|\begin{array}{ccc}x+a+b+c& b& c\\ x+a+b+c& x+b& c\\ x+a+b+c& b& x+c\end{array}\right|=0$

Now, $R_3\to R_3-R_1$ and $R_2\to R_2-R_1$, we get

$\Delta =\left|\begin{array}{ccc}x+a+b+c& b& c\\ 0& x& 0\\ 0& 0& x\end{array}\right|=0$

$\Delta =x\left(x\right(x+a+b+c\left)\right)=0$

It's given that $x\ne 0$

$\therefore x=-(a+b+c)$