$U = \left[ {\begin{array}{*{20}{c}}1&1&0\2&3&4\end{array}} \right]\,\,\;\quad V = \left[ {\begin{array}{*{20}{c}}3\1\end{array}} \right]\,\,\;\quad W = \left[ {\begin{array}{*{20}{c}}{0.6}&{0.5}\{0.4}&{0.5}\end{array}} \right]\;\,\,\;X = \left[ {\begin{array}{*{20}{c}}3&5\4&6\end{array}} \right]\quad \,\, \ Y = \left[ {\begin{array}{*{20}{c}}1&3&2\0&5&6\0&0&1\end{array}} \right]\;\quad $ det (W) =

Question

BlackTom AI · Accepted Answer

To compute the determinant of the 2x2 matrix W = [[0.6, 0.5], [0.4, 0.5]], we apply the standard formula det(W) = (a*d) − (b*c), where a = 0.6, b = 0.5, c = 0.4, d = 0.5. Multiplying the diagonal elements gives a*d = 0.6 × 0.5 = 0.30. Next, multiplying the off-diagonal elements gives b*c = 0.5 × 0.4 = 0.20. Subtracting these results yields det(W) = 0.30 − 0.20 = 0.10. In decimal form, this is 0.10, which is the precise determinant value. Note that this aligns with the general expectation that the determinant of a 2×2 matrix reflects area scaling (or orientation) and is independent of any other matrices provided in the prompt unless they are involved in a larger context. Since no alternative answer choices are shown here, I’m focusing on the direct computation for det(W).

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