Question

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

Answer 1

Let $\Delta=\begin{vmatrix}\ x^2+x&x+1&x-2\\2x^2+3x-1& 3x& 3x-3\\x^2+2x+3& 2x-1& 2x-1 \end{vmatrix}$
Applying $R_2\rightarrow R_2 - (R_1 + R_3 )$, we get
$\Delta=\begin{vmatrix}\ x^2+x&x+1&x-2\\-4& 0& 0\\x^2+2x+3& 2x-1& 2x-1 \end{vmatrix}$
Applying $R_1 \rightarrow R_1+\frac{x^2}{4} R_2$
and $R_3\rightarrow R_3+\frac{x_2}{4}R_2, $ we get
$\Delta=\begin{vmatrix}\ x&x+1&x-2\\-4& 0& 0\\2x+3& 2x-1& 2x-1 \end{vmatrix}$
Applying $R_3\rightarrow R_3-2R_1=\begin{vmatrix}\ x+0&x+1&x-2\\-4& 0& 0\\3& -3& 3 \end{vmatrix}$
$=\begin{vmatrix}\ x&x&x\\-4& 0& 0\\3& -3& 3 \end{vmatrix}+\begin{vmatrix}\ 0&1&-2\\-4& 0& 0\\3& -3& 3 \end{vmatrix}$
$=x\begin{vmatrix}\ 1&1&1\\-4& 0& 0\\3& -3& 3 \end{vmatrix}+\begin{vmatrix}\ 0&1&-2\\-4& 0& 0\\3& -3& 3 \end{vmatrix}$
$\Rightarrow \Delta=Ax+B$
where, $A=\begin{vmatrix}\ 1&1&1\\-4& 0& 0\\3& -3& 3 \end{vmatrix}$ and $B=\begin{vmatrix}\ 0&1&-2\\-4& 0& 0\\3& -3& 3 \end{vmatrix}$