Question

The value of $a$ for which system of equations, $a^{3}x+{(a+1)}^{3}y+{(a+2)}^{3}z=0,ax+(a+1)y+(a+2)z=0,x+y+z=0$ has non zero solutions is :

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $-1$

Find value of a for which system of equation

$a^{3}x+(a+1{)}^{3}y+(a+2{)}^{3}z=0,ax+(a+1)y+(a+2)z=0,$

$x+y+z=0$ has non-zero solution

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

$R_1\to R_1-R_3,R_2\to R_2-R_3$

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

expand along $R_3$ is

$\perp [-a^3+(a+2{)}^3+2(a+1{)}^3-2(a+2{)}^3]=0$

$\Rightarrow [-a^3-(a+2{)}^3+2(a+1{)}^3]=0$

$\Rightarrow [-a^3-(a^3+8+6a^2+10q)+2a^3+2+6a^2+6a=0]$

$[-a^3-a^3-86a^2-12a+2a^3+2+6a^2+6A]=0$

$[-a^3-q^3-8-6a^{2}12-a^3+2+6a^2+6a0]$

$=-6a-0$

$-1$