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# Without expanding show that $\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2 +3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = xA +B$ Also find A and B.

$(A)\; A = \begin{vmatrix} 1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} \;and \; B= \begin{vmatrix} 0 & 1 & -2\\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$
$(B)\; A = \begin{vmatrix} 1 & -1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} \;and \; B= \begin{vmatrix} 0 & 1 & 2\\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$
$(C)\; A = \begin{vmatrix} 1 & 1 & -1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} \;and \; B= \begin{vmatrix} 0 & -1 & -2\\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$
$(D)\; A = \begin{vmatrix} 1 & 1 & 1 \\ -4 & 0 & 0 \\ -3 & 3 & 3 \end{vmatrix} \;and \; B= \begin{vmatrix} 0 & 1 & -2\\ -4 & 0 & 0 \\ -3 & 3 & 3 \end{vmatrix}$

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A)
$R_2 \to R_2-R_1-R_3$
$= \begin{vmatrix} x^2+x & x+1 & x-2 \\ -4 & 0 & 0 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix}$
$R_1 \to R_1 +\frac{x^2}{4} R_2 ,\; \; \; R_3 \to R_3 + \frac{x^2}{4}R_2$
$= \begin{vmatrix} x & x+1 & x-2 \\ -4 & 0 & 0 \\ 2x-3 & 2x-1 & 2x-1 \end{vmatrix}$
$R_3 \to R_3 -2R_1$
$= \begin{vmatrix} x & x+1 & x-2 \\ -4 & 0 & 0 \\ 3 & -3& 3 \end{vmatrix}$
$= \begin{vmatrix} x & x& x \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} +\begin{vmatrix} 0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$
$= x \begin{vmatrix} 1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} + \begin{vmatrix} 0 & 1 & -2\\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$
$= xA + B$ Hence Proved
$\therefore A = \begin{vmatrix} 1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix} \;and \; B= \begin{vmatrix} 0 & 1 & -2\\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{vmatrix}$\$