[Solved] What is the output of this code?import numpy as np from scipy.linalg import lu A = np.array([[4, 3], [6, 3]]) P, L, U = lu(A) print(np.round(L @ U, 2)) | SciPy

SciPy - Linear Algebra (scipy.linalg)

What is the output of this code?

import numpy as np
from scipy.linalg import lu
A = np.array([[4, 3], [6, 3]])
P, L, U = lu(A)
print(np.round(L @ U, 2))

A[[6. 3.] [4. 3.]]

B[[1. 0.] [0. 1.]]

C[[3. 4.] [3. 6.]]

D[[0. 1.] [1. 0.]]

Step-by-Step Solution

Solution:

Step 1: Understand LU decomposition and reconstruction
LU decomposition factors A into P, L, U such that P@A = L@U. Multiplying L and U reconstructs P@A, not A directly.
Step 2: Check output of L@U
Since P is a permutation matrix, L@U equals P@A. For this matrix, P swaps rows, so L@U equals the permuted A matrix, which is [[6,3],[4,3]].
Final Answer:
[[6. 3.] [4. 3.]] -> Option A
Quick Check:
L@U = P@A = permuted A matrix [OK]

Quick Trick: L times U equals permuted A matrix, not original A [OK]

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What is the output of this code?