Question

If the value of determinant $\Delta =\left|\begin{array}{ccc}{a}^2+a^3-3& a^2& a^3\\ 77& 16& 64\\ a^4+a^6-3& a^4& a^6\end{array}\right|$ is zero. Then the number of different value(s) of $a$ is

Answer 1

Given : $\Delta =\left|\begin{array}{ccc}{a}^2+a^3-3& a^2& a^3\\ 77& 16& 64\\ a^4+a^6-3& a^4& a^6\end{array}\right|$
Applying $C_1\to C_1-(C_2+C_3)$, we get

$\Delta =\left|\begin{array}{ccc}-3& a^2& a^3\\ -3& 4^2& 4^3\\ -3& (a^2{)}^2& (a^2{)}^3\end{array}\right|=0$
Taking $(-3)$ common from $C_1,$ we get
$\Rightarrow (-3)\left|\begin{array}{ccc}1& a^2& a^3\\ 1& 4^2& 4^3\\ 1& (a^2{)}^2& (a^2{)}^3\end{array}\right|=0$
$\Rightarrow -3(a-4)(4-a^2)(a^2-a)(4a+4a^2+a^3)=0\because \left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|=(a-b)(b-c)(c-a)(ab+bc+ca)\Rightarrow a^2(a-4)(4-a^2)(a-1)(a^2+4a+4)=0\Rightarrow a^2(a-1)(2-a)(2+a)(a+2{)}^2(a-4)=0$
So, value(s) of $a=-2,0,1,2,4$