GATE DA 2024

\(\boldsymbol{M} = \boldsymbol{u}\boldsymbol{u}^{T}\) is a symmetric \(5 \times 5\) matrix of rank 1, since every column is a multiple of \(\boldsymbol{u}\). Moreover

\[ \boldsymbol{M}\boldsymbol{u} = \boldsymbol{u}(\boldsymbol{u}^{T}\boldsymbol{u}) = \|\boldsymbol{u}\|^{2}\,\boldsymbol{u}, \qquad \|\boldsymbol{u}\|^{2} = 1 + 4 + 9 + 16 + 25 = 55 \]

while every vector orthogonal to \(\boldsymbol{u}\) is sent to \(\boldsymbol{0}\). So the eigenvalues are 55 and 0 (four times). For a symmetric matrix the singular values are the absolute values of the eigenvalues, here 55, 0, 0, 0, 0.

Hence \(\sum_{i=1}^{5} \sigma_{i} = 55\), which is also the trace of \(\boldsymbol{M}\).

Answer: 55

Information gain is the entropy of the Target minus the weighted entropy left after splitting on Pitch (logarithms to base 2).

• All 10 games: 4 Green and 6 Blue, so \(H(C) = -0.4\log_{2}0.4 - 0.6\log_{2}0.6 \approx 0.971\).
• \(S\) pitches (games 1, 2, 4, 7, 10): 3 Green and 2 Blue, the same 0.6 / 0.4 split with the colours swapped, so the entropy is again \(\approx 0.971\).
• \(F\) pitches (games 3, 5, 6, 8, 9): 1 Green and 4 Blue, so the entropy is \(-0.2\log_{2}0.2 - 0.8\log_{2}0.8 \approx 0.722\).

Each branch holds half of the games, so

\[ \text{Gain} = 0.971 - \left(\tfrac{1}{2} \times 0.971 + \tfrac{1}{2} \times 0.722\right) = \tfrac{1}{2}(0.971 - 0.722) \approx 0.1245 \]

which is 0.12 to two decimal places (the key accepts 0.12 to 0.13).

Answer: 0.12

Distances from the diamond at (2, 1):

• circle (1, 1): 1
• circle (1, 2) and square (3, 2): \(\sqrt{2} \approx 1.41\) each
• square (2, 3): 2
• square (3, 3): \(\sqrt{5} \approx 2.24\)

Now try the odd values of \(k\) in turn:

• \(k = 1\): the nearest point is a circle, so the label is circle.
• \(k = 3\): two circles and one square; the tie at \(\sqrt{2}\) does not matter because both tied points are among the three. The label is circle.
• \(k = 5\): all five points vote, 3 squares to 2 circles, so the label is square.

Answer: 5

A Bayesian network factorises the joint probability into one factor per node, each conditioned on that node's parents. Here U has no parents, V and W each have parent U, and Z has parents V and W:

\[ P(U, V, W, Z) = P(U)\,P(V|U)\,P(W|U)\,P(Z|V, W) \]

Reading the tables for \(U = 1, V = 1, W = 1, Z = 1\):

• \(P(U = 1) = 0.5\)
• \(P(V = 1|U = 1) = 0.5\)
• \(P(W = 1|U = 1) = 1\)
• \(P(Z = 1|V = 1, W = 1) = 0.5\)

So the joint probability is \(0.5 \times 0.5 \times 1 \times 0.5 = 0.125\).

Answer: 0.125

Use \(\operatorname{Cov}(X, Y) = E[XY] - E[X]\,E[Y]\). The four outcomes HH, HT, TH, TT are equally likely.

• \(X = 1\) only for HH, so \(E[X] = \frac{1}{4}\).
• \(Y = 1\) for every outcome except TT, so \(E[Y] = \frac{3}{4}\).
• Whenever \(X = 1\), also \(Y = 1\), so \(XY = X\) and \(E[XY] = \frac{1}{4}\).

\[ \operatorname{Cov}(X, Y) = \frac{1}{4} - \frac{1}{4} \cdot \frac{3}{4} = \frac{1}{16} = 0.0625 \]

The covariance is positive because two heads guarantee at least one head. As 0.0625 lies exactly halfway, the key accepts both 0.062 and 0.063.

Answer: 0.063