$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 $ The upper triangular matrix is:

Question

BlackTom AI · Accepted Answer

Let's break down what an upper triangular matrix means for this question. An upper triangular matrix must be square (same number of rows and columns) and have all entries below the main diagonal equal to zero.

Option U: The matrix U is 2 by 3 (not square). Since it isn’t square, it cannot be classified as upper triangular by the standard definition, which requires a square shape.

Option V: The matrix V is 2 by 1 (also not square). Its shape disqualifies it from being upper triangular for the same reason as U.

Option W: The matrix W is 2 by 2, so it’s square, but looking at its entries, the element below the main diagonal is 0.4, not 0. This means W has a nonzero entry below the diagonal, so it is not upper triangular.

Option X: The matrix X is 2 by 2 and square, yet the element in the position (2,1) is 4, which is below the diagonal and not zero. Therefore, X is not upper triangular.

Option Y: The matrix Y is 3 by 3 and square. All entries below the main diagonal are zero (positions (2,1) and (3,1), (3,2) are 0). This satisfies the defining condition for an upper triangular matrix, so Y is upper triangular.

In summary, the only matrix among the given options that meets both requirements—being square and having zeros below the main diagonal—is Y, making it the upper triangular matrix.

MCD1550 / MCD2140 - T3 - 2024 11.6 Week 11: In Class - Part B

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