-5
-2
2
5
Correct answer is D
\(P=\left[\begin{array}{cc}x+3 & x+2\\
x+1 & x-1\end{array}\right]\) evaluate x if |P| = -10
(x+3)(x-1) - {(x+1)(x+2)} = -10
x2 - x + 3x - 3 - {x2 + 2x + x + 2} = -10
x2 + 2x - 3 - {x2 + 3x + 2} = -10
-x - 5 = -10
-5 + 10 = x
5 = x
∴x = 5
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