Question

[tex] \left |\begin{array}{ccc}a+x&a-x&a-x\\a-x&a+x&a-x\\a-x&a-x&a+x\end{array}\right | = 0 are x=0 and x=3a,Prove roots of the equation. [\tex]

Answer 1

Step-by-step explanation:

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

$R_1 - > R_1+ R_2 + R_3$

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

$\left |\begin{array}{ccc}3a-x&3a-x&3a-x\\a-x&a+x&a-x\\a-x&a-x&a+x\end{array}\right | = 0$

Take (3a-x) common from R1

$(3a-x)\left |\begin{array}{ccc}1&1&1\\a-x&a+x&a-x\\a-x&a-x&a+x\end{array}\right | = 0$

$C_1 - > C_1- C_3 \\ \\C_2 - > C_2- C_3$

$(3a-x)\left |\begin{array}{ccc}1-1&1-1&1\\a-x-a+x&a+x-a+x&a-x\\a-x-a-x&a-x-a-x&a+x\end{array}\right | = 0$

$(3a-x)\left |\begin{array}{ccc}0&0&1\\0&2x&a-x\\-2x&2x&a+x\end{array}\right | = 0$

Now expand the determinant along R1

$(3a-x)(4x)=0\\\\3a-x=0\\\\x=3a\\\\or\\\\4x=0\\\\x=0$

Hence proved

Hope it helps you.