np.linalg.inv() for matrix inverse in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of matrix inverse calculation

What is the output of this code snippet using np.linalg.inv()?

NumPy

import numpy as np
A = np.array([[4, 7], [2, 6]])
inv_A = np.linalg.inv(A)
print(inv_A)

A

[[ 1.5 -1.7]
 [-0.2  0.4]]

B

[[ 0.6  0.7]
 [ 0.2  0.4]]

C

[[ 0.6 -0.7]
 [ 0.2 -0.4]]

D

[[ 0.6 -0.7]
 [-0.2  0.4]]

💡 Hint

Recall the formula for 2x2 matrix inverse: 1/det * [[d, -b], [-c, a]]

The determinant of A is (4*6 - 7*2) = 10. The inverse is 1/10 * [[6, -7], [-2, 4]] which matches option D.

❓ data_output

intermediate

Shape of inverse matrix

Given a square matrix B of shape (3, 3), what is the shape of np.linalg.inv(B)?

A (9,)

B (3,)

C (3, 3)

D (1, 3)

💡 Hint

The inverse of a square matrix has the same shape as the original matrix.

The inverse matrix has the same dimensions as the original square matrix, so shape remains (3, 3).

🔧 Debug

advanced

Error raised by singular matrix inverse

What error does this code raise when trying to invert a singular matrix?

NumPy

import numpy as np
C = np.array([[1, 2], [2, 4]])
inv_C = np.linalg.inv(C)

A TypeError: unsupported operand type(s)

B numpy.linalg.LinAlgError: Singular matrix

C ValueError: shapes not aligned

D IndexError: index out of bounds

💡 Hint

A singular matrix has determinant zero and cannot be inverted.

The matrix C is singular because its rows are linearly dependent, so np.linalg.inv raises LinAlgError.

🧠 Conceptual

advanced

Matrix inverse property check

If D is an invertible matrix, which expression correctly checks if inv_D is the inverse of D?

A np.allclose(np.dot(D, inv_D), np.eye(D.shape[0]))

B np.dot(D, inv_D) == np.eye(D.shape[0])

C np.array_equal(np.dot(D, inv_D), np.eye(D.shape[0]))

D np.isclose(np.dot(D, inv_D), np.eye(D.shape[0]))

💡 Hint

Use a function that allows for floating point tolerance when comparing arrays.

np.allclose checks if two arrays are element-wise equal within a tolerance, suitable for floating point results.

🚀 Application

expert

Inverse of a matrix product

Given two invertible matrices E and F, which expression correctly computes the inverse of their product E @ F?

A np.linalg.inv(F) @ np.linalg.inv(E)

B np.linalg.inv(E) @ np.linalg.inv(F)

C np.linalg.inv(E @ F)

D np.linalg.inv(F) + np.linalg.inv(E)

💡 Hint

Recall the rule: (AB)^-1 = B^-1 A^-1

The inverse of a product is the product of the inverses in reverse order.