You have a sparse matrix A of size 20000x20000 representing connections in a social network, and a vector b of known values. Which method is most efficient to solve Ax = b?

hard📝 Application Q8 of 15

SciPy - Sparse Linear Algebra

You have a sparse matrix A of size 20000x20000 representing connections in a social network, and a vector b of known values. Which method is most efficient to solve Ax = b?

AConvert A to a dense matrix and use numpy.linalg.solve

BUse iterative solvers like conjugate gradient without preconditioning

CUse <code>spsolve</code> with A in CSR or CSC sparse format

DManually invert A and multiply by b

Step-by-Step Solution

Solution:

Step 1: Consider matrix size and sparsity
A is large and sparse, so dense methods are inefficient.
Step 2: Choose solver
spsolve is optimized for sparse matrices in CSR or CSC format and uses direct methods efficiently.
Step 3: Evaluate other options
Iterative solvers may be slower or require tuning; manual inversion is computationally expensive and unstable.
Final Answer:
Use spsolve with A in CSR or CSC sparse format -> Option C
Quick Check:
Check if A is sparse and large, prefer sparse direct solvers [OK]

Quick Trick: Use spsolve with sparse formats for large sparse systems [OK]

Common Mistakes:

Converting sparse to dense unnecessarily
Using manual inversion for large matrices
Ignoring sparse matrix formats for spsolve

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You have a sparse matrix A of size 20000x20000 representing connections in a social network, and a vector b of known values. Which method is most efficient to solve Ax = b?

Step 1: Consider matrix size and sparsity

Step 2: Choose solver

Step 3: Evaluate other options

Final Answer:

Quick Check:

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