Question

If the following system of equations is consistent,
${(a+1)}^{3}x+{(a+2)}^{3}y={(a+3)}^3,(a+1)x+(a+2)y=a+3,x+y=1$, then find the value of $a$.

Answer 1

Given system of equations is consistent.
So $D=0$
Here, $D=\left|\begin{array}{ccc}(a+1{)}^3& (a+2{)}^3& -(a+3{)}^3\\ a+1& a+2& -(a+3)\\ 1& 1& -1\end{array}\right|$
$\Rightarrow \left|\begin{array}{ccc}(a+1{)}^3& (a+2{)}^3& -(a+3{)}^3\\ a+1& a+2& -(a+3)\\ 1& 1& -1\end{array}\right|=0$
$\Rightarrow -\left|\begin{array}{ccc}(a+1{)}^3& (a+2{)}^3& (a+3{)}^3\\ a+1& a+2& (a+3)\\ 1& 1& 1\end{array}\right|=0$
$C_1\to C_1-C_2;C_2\to C_2-C_3$
$\Rightarrow \left|\begin{array}{ccc}-(3a^2+9a+7)& -(3a^2+15a+19)& (a+3{)}^3\\ -1& -1& a+3\\ 0& 0& 1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}(3a^2+9a+7)& (3a^2+15a+19)& (a+3{)}^3\\ 1& 1& a+3\\ 0& 0& 1\end{array}\right|=0$
$C_1\to C_1-C_2$
$\Rightarrow \left|\begin{array}{ccc}-6a-12& (3a^2+15a+19)& (a+3{)}^3\\ 0& 1& a+3\\ 0& 0& 1\end{array}\right|=0$
$\Rightarrow 6a+12=0$
$\Rightarrow a=-2$
So $|a|=2$