Question

Prove that the roots of the equation
| a+x a-x a-x |
| a-x a+x a-x | = 0 are x=0 and x=3a
| a-x a-x a+x |

Answer 1

Step-by-step explanation:

Given:

$=\begin{vmatrix}a+x&a-x&a-x \\ a-x&a+x&a-x \\ a-x&a-x&a+x\end{vmatrix}=0$

We have to prove values of x are 0 and 3a.

$R_1=>R_1+R_2+R_3$

$=\begin{vmatrix}3a-x&a-x&a-x \\ 3a-x&a+x&a-x \\ 3a-x&a-x &a+x\end{vmatrix}=0$

After taking 3a-x from coloumn first, we get :

$=(3a-x)\begin{vmatrix}1&a-x&a-x \\ 1&a+x&a-x \\1&a-x&a+x\end{vmatrix}=0$

Taking the following operations:

$R_2=>R_2-R_1\ and \ R_3=>R_3-R_1$

$=(3a-x)\begin{vmatrix}1&a-x&a-x\\ 0&2x& 0\\0&0&2x\end{vmatrix}=0$

Expanding along column first , We get:

$=(3a-x)\begin{vmatrix}2x &0\\ 0 &2x \end{vmatrix}=0$

Upon solving the determinant, We get:

$(3a-x)(4x^2)=0$

$3a-x=0 \ and\ 4x^2=0$

x=3a and x=0 [ upon solving above equation]