[math: U=[110234]V=[31]W=[0.60.50.40.5]X=[3546]Y=[132056001]]U = \left[ {\begin{array}{*{20}{c}}1&1&0\2&3&4\end{array}} ight]\,\,\;\quad V = \left[ {\begin{array}{*{20}{c}}3\1\end{array}} ight]\,\,\;\quad W = \left[ {\begin{array}{*{20}{c}}{0.6}&{0.5}\{0.4}&{0.5}\end{array}} ight]\;\,\,\;X = \left[ {\begin{array}{*{20}{c}}3&5\4&6\end{array}} ight]\quad \,\, \ Y = \left[ {\begin{array}{*{20}{c}}1&3&2\0&5&6\0&0&1\end{array}} ight]\;\quad det (W) =

Question

[math: U=[110234]V=[31]W=[0.60.50.40.5]X=[3546]Y=[132056001]]U = \left[ {\begin{array}{*{20}{c}}1&1&0\2&3&4\end{array}} ight]\,\,\;\quad V = \left[ {\begin{array}{*{20}{c}}3\1\end{array}} ight]\,\,\;\quad W = \left[ {\begin{array}{*{20}{c}}{0.6}&{0.5}\{0.4}&{0.5}\end{array}} ight]\;\,\,\;X = \left[ {\begin{array}{*{20}{c}}3&5\4&6\end{array}} ight]\quad \,\, \ Y = \left[ {\begin{array}{*{20}{c}}1&3&2\0&5&6\0&0&1\end{array}} ight]\;\quad det (W) =

BlackTom AI · Accepted Answer

To determine the determinant of the matrix W, recall the standard formula for a 2x2 matrix W = [[a, b], [c, d]]: det(W) = ad − bc. Here, W is given as [[0.6, 0.5], [0.4, 0.5]]. Identify the entries: a = 0.6, b = 0.5, c = 0.4, d = 0.5. Compute the products: ad = 0.6 × 0.5 = 0.30 and bc = 0.5 × 0.4 = 0.20. Subtract to obtain det(W) = 0.30 − 0.20 = 0.10. Therefore, the determinant evaluates to 0.1. Note that the given answer format lists a single option “c. 0.1”; since no other options are provided to contrast, the calculation above matches the numerical value of the determinant. If one were to consider alternative distractors, they would typically involve misplacing a term or performing the wrong operation (e.g., summing instead of subtracting, or using incorrect pairings), but with the correct application of ad − bc, the result is 0.1.

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