Solve the equation, if ∣∣ ∣∣x+axxxx+axxxx+a∣∣ ∣∣=0,a≠0,
Given, ∣∣
∣∣x+axxxx+axxxx+a∣∣
∣∣=0⇒∣∣
∣∣3x+a3x+a3x+axx+axxxx+a∣∣
∣∣=0
(using R1→R1+R2+R3)
⇒(3x+a)∣∣
∣∣111xx+axxxx+a∣∣
∣∣=0
(Taking out (3x+a) from R1)
⇒(3x+a)∣∣
∣∣100xa0xa0∣∣
∣∣=0, (using C2→C2−C1 and C3→C3−C1)
Expanding corresponding to R1 (first row), we get
(3x+a)[1(a×a−0)]=0⇒a2(3x+a)=0
But a≠0 (given). Therefore, 3x+a=0⇒x=−a/3