Question

Prove that: $\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|=(a-1{)}^C$
then $C=$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\Delta =\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|$

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

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

$=1((1-a^2)(2-2a)-(3-a^2-2a)(1-a))$

$=(1-a)((1+a)(2-2a)+(a-1)(a+3))$

$={(1-a)}^2(2+2a-a-3)={(1-a)}^2(a-1)$

$={(a-1)}^3$