If $|\begin{array}{ccc} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{array}| = xA + B$ then

Moderate

A

$A = |\begin{array}{ccc} 1 & 1 & 1 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}|$
B

$A = |\begin{array}{ccc} 0 & 1 & 2 \\ 1 & - 2 & 3 \\ - 4 & 0 & 0 \end{array}|$
C

$B = |\begin{array}{ccc} 1 & 1 & - 2 \\ - 3 & - 2 & 3 \\ 4 & 0 & 1 \end{array}|$
D

$B = |\begin{array}{ccc} 0 & 1 & - 2 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}|$

Solution

Let $Δ = |\begin{array}{ccc} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{array}|$
Applying $R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - R_{1}$ , we get
$\begin{matrix} Δ = |\begin{array}{ccc} x^{2} + x & x + 1 & x - 2 \\ x - 1 & x - 2 & x + 1 \\ x + 3 & x - 2 & x + 1 \end{array}| \\ = |\begin{array}{ccc} x^{2} & x + 1 & x - 2 \\ 0 & x - 2 & x + 1 \\ 0 & x - 2 & x + 1 \end{array}| + |\begin{array}{ccc} x & x + 1 & x - 2 \\ x - 1 & x - 2 & x + 1 \\ x + 3 & x - 2 & x + 1 \end{array}| \\ = 0 + |\begin{array}{ccc} x & x + 1 & x - 2 \\ x - 1 & x - 2 & x + 1 \\ x + 3 & x - 2 & x + 1 \end{array}| \end{matrix}$
Applying $R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{2}$ , we get
$Δ = |\begin{array}{ccc} x & x + 1 & x - 2 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}| \begin{matrix} = |\begin{array}{ccc} x & x & x \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}| + |\begin{array}{ccc} 0 & 1 & - 2 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}| \\ = |\begin{array}{ccc} 1 & 1 & 1 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}| x + |\begin{array}{ccc} 0 & 1 & - 2 \\ - 1 & - 3 & 3 \\ 4 & 0 & 0 \end{array}| \end{matrix}$

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