$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 $ X –1 =

Question

BlackTom AI · Accepted Answer

The problem asks for X^{-1}, i.e., the inverse of the 2x2 matrix X.
First, identify X: X = [ [3, 5], [4, 6] ].
For a 2x2 matrix [ [a, b], [c, d] ], the inverse exists if det(X) = ad - bc ≠ 0, and is given by (1/det) * [ [d, -b], [-c, a] ].
Compute the determinant: det(X) = (3)(6) - (5)(4) = 18 - 20 = -2. Since det(X) ≠ 0, the inverse exists.
Next, compute the adjugate (the matrix of cofactors transposed): [ [d, -b], [-c, a] ] = [ [6, -5], [-4, 3] ].
Multiply by 1/det(X) = 1/(-2) = -1/2 to obtain X^{-1} = (-1/2) * [ [6, -5], [-4, 3] ].
Carrying out the scalar multiplication:
- (-1/2) * 6 = -3,
- (-1/2) * (-5) = 2.5,
- (-1/2) * (-4) = 2,
- (-1/2) * 3 = -1.5.
Thus X^{-1} = [ [ -3, 2.5 ], [ 2, -1.5 ] ].
The provided answer (b) matches this result exactly, confirming it is correct.
Note: The prompt shows an answer option but provides no full list of answer choices. In practice, when evaluating, you would compare the computed X^{-1} with each given option to determine correctness; here the computed inverse aligns with the shown option.

MCD1550 / MCD2140 - T1 - 2025 TEST 2 (Week 6)

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