[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, 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, 2))

A[[1. 0. ] [0.67 1. ]]

B[[4. 3. ] [6. 3. ]]

C[[1.5 0. ] [0. 1. ]]

D[[0.67 1. ] [1. 0. ]]

Step-by-Step Solution

Solution:

Step 1: Perform LU decomposition on matrix A
Using scipy.linalg.lu on A = [[4,3],[6,3]] gives matrices P, L, U where L is lower triangular with 1s on diagonal and multipliers below.
Step 2: Calculate L matrix values
L = [[1, 0], [6/4, 1]] = [[1, 0], [1.5, 1]]. But scipy.linalg.lu applies row permutations, so actual L is [[1,0],[0.67,1]] after permutation.
Final Answer:
[[1. 0. ] [0.67 1. ]] -> Option A
Quick Check:
L matrix lower triangular with 1 diagonal = [[1,0],[0.67,1]] [OK]

Quick Trick: L is lower triangular with 1s on diagonal [OK]

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What is the output of this code?