Question

If $\Delta =\left|\begin{array}{ccc}x-3& 2x+1& 2\\ 3x+2& x+2& 1\\ 5x+1& 5x+4& 5\end{array}\right|$, then $\Delta$ is

• A. multiple of $x^2$
• B. $15$
• C. a multiple of $x$
• D. $-15$

Answer 1

The correct option is D $-15$

$\left|\begin{array}{ccc}x-3& 2x+1& 2\\ 3x+2& x+2& 1\\ 5x+1& 5x+4& 5\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}2(x-3)& 2(2x+1)& 2\left(2\right)\\ 3x+2& x+2& 1\\ 5x+1& 5x+4& 5\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}2x-6& 4x+2& 4\\ 3x+2& x+2& 1\\ 5x+1& 5x+4& 5\end{array}\right|$

Apply $R_1\to R_1+R_2-R_3$

$=\frac{1}{2}\left|\begin{array}{ccc}-5& 0& 0\\ 3x+2& x+2& 1\\ 5x+1& 5x+4& 5\end{array}\right|$

$=\frac{1}{2}\times (-5)\left\{5\right(x+2)-1(5x+4\left)\right\}$

$=\frac{-5}{2}\times 6$

$=-15$