Question

Without expanding a determinant at any stage, show that $\left|\begin{array}{ccc}{x}^2+x& x+1& x-2\\ 2x^2+3x-1& 3x& 3x-3\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|=xA+B$ where $A$ and $B$ are determinants of order $3$ not involving $x$.

Answer 1

Let $\Delta =\left|\begin{array}{ccc}{x}^2+x& x+1& x-2\\ 2x^2+3x-1& 3x& 3x-3\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|$
Applying $R_2\to R_2-(R_1+R_3)$, we get
$\Delta =\left|\begin{array}{ccc}{x}^2+x& x+1& x-2\\ -4& 0& 0\\ x^2+2x+3& 2x-1& 2x-1\end{array}\right|$
Applying $R_1\to R_1+\frac{x^2}{4}{R}_2$
and $R_3\to R_3+\frac{x_2}{4}{R}_2,$ we get
$\Delta =\left|\begin{array}{ccc}x& x+1& x-2\\ -4& 0& 0\\ 2x+3& 2x-1& 2x-1\end{array}\right|$
Applying $R_3\to R_3-2R_1=\left|\begin{array}{ccc}x+0& x+1& x-2\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|$
$=\left|\begin{array}{ccc}x& x& x\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|+\left|\begin{array}{ccc}0& 1& -2\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|$
$=x\left|\begin{array}{ccc}1& 1& 1\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|+\left|\begin{array}{ccc}0& 1& -2\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|$
$\Rightarrow \Delta =Ax+B$
where, $A=\left|\begin{array}{ccc}1& 1& 1\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|$ and $B=\left|\begin{array}{ccc}0& 1& -2\\ -4& 0& 0\\ 3& -3& 3\end{array}\right|$