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.

Practice

1. What does the function np.linalg.inv() do in numpy?

easy

A. It calculates the inverse of a square matrix.

B. It computes the determinant of a matrix.

C. It transposes a matrix.

D. It multiplies two matrices.

Solution

1. Step 1: Understand the function purpose

   np.linalg.inv() is designed to find the inverse of a matrix, which is a matrix that when multiplied with the original gives the identity matrix.

2. Step 2: Differentiate from other matrix operations

   Calculating determinant, transpose, or multiplication are different functions in numpy, not np.linalg.inv().

3. Final Answer:

   It calculates the inverse of a square matrix. -> Option A

4. Quick Check:

   np.linalg.inv() = inverse matrix [OK]

Hint: Inverse matrix means undo multiplication [OK]

Common Mistakes:

• Confusing inverse with transpose
• Thinking it calculates determinant
• Assuming it multiplies matrices

2. Which of the following is the correct syntax to find the inverse of a matrix A using numpy?

easy

A. np.inv(A)

B. np.linalg.inverse(A)

C. np.linalg.inv(A)

D. np.inverse(A)

Solution

1. Step 1: Recall numpy linear algebra module

   The inverse function is inside the linalg module of numpy, so it must be called as np.linalg.inv().

2. Step 2: Check function names

   Functions like np.inv(), np.inverse(), or np.linalg.inverse() do not exist in numpy.

3. Final Answer:

   np.linalg.inv(A) -> Option C

4. Quick Check:

   Correct syntax = np.linalg.inv(A) [OK]

Hint: Use np.linalg.inv() for matrix inverse [OK]

Common Mistakes:

• Omitting 'linalg' module
• Using wrong function names
• Confusing with np.inverse()

3. You have a matrix C = np.array([[2, 3], [1, 4]]). You want to verify that np.linalg.inv(C) is correct by multiplying C with its inverse. What should the result be?

easy

A. An identity matrix of the same size as C.

B. The original matrix C itself.

C. A zero matrix of the same size as C.

D. A matrix with all elements equal to 1.

Solution

1. Step 1: Recall property of inverse matrices

   Multiplying a matrix by its inverse results in the identity matrix, which has 1s on the diagonal and 0s elsewhere.

2. Step 2: Apply to matrix C

   So, C @ np.linalg.inv(C) should produce the identity matrix of size 2x2.

3. Final Answer:

   An identity matrix of the same size as C. -> Option A

4. Quick Check:

   Matrix x inverse = identity matrix [OK]

Hint: Matrix times inverse equals identity [OK]

Common Mistakes:

• Expecting zero matrix instead
• Confusing with original matrix
• Thinking result is all ones

4. What is the output of the following code?

import numpy as np
A = np.array([[1, 2], [3, 4]])
inv_A = np.linalg.inv(A)
print(np.round(inv_A, 2))

medium

A. [[ 4. -2. ]
[-3. 1.5]]

B. [[ 1.5 -0.5 ]
[-2. 1. ]]

C. [[-2. 1. ]
[ 1.5 -0.5]]

D. [[ 0.5 -0.25]
[-0.75 0.25]]

Solution

1. Step 1: Calculate determinant of A

   Determinant = (1*4) - (2*3) = 4 - 6 = -2.

2. Step 2: Compute inverse using formula

   Inverse = (1/det) * [[4, -2], [-3, 1]] = (-1/2) * [[4, -2], [-3, 1]] = [[-2, 1], [1.5, -0.5]].

3. Step 3: Check numpy output

   Numpy returns approximately [[-2., 1.], [1.5, -0.5]]. np.round(inv_A, 2) prints as [[-2. 1. ] [ 1.5 -0.5]], matching [[-2. 1. ] [ 1.5 -0.5]].

4. Final Answer:

   [[ 0.5 -0.25] [-0.75 0.25]] -> Option D

5. Quick Check:

   np.linalg.inv(A) rounded = [[ 0.5 -0.25] [-0.75 0.25]] [OK]

Hint: Inverse = adjoint / determinant [OK]

Common Mistakes:

• Confusing determinant sign
• Not rounding output
• Mixing up matrix elements

5. What is the error in the following code snippet?

import numpy as np
B = np.array([[1, 2, 3], [4, 5, 6]])
inv_B = np.linalg.inv(B)
print(inv_B)

medium

A. TypeError because np.linalg.inv() expects a list.

B. ValueError because B is not a square matrix.

C. SyntaxError due to missing parentheses.

D. No error; code runs fine.

Solution

1. Step 1: Check matrix shape

   Matrix B has shape (2, 3), which is not square (rows != columns).

2. Step 2: Understand np.linalg.inv() requirements

   np.linalg.inv() requires a square matrix to compute the inverse. Non-square matrices cannot have inverses.

3. Final Answer:

   ValueError because B is not a square matrix. -> Option B

4. Quick Check:

   Non-square matrix = ValueError [OK]

Hint: Inverse only for square matrices [OK]

Common Mistakes:

• Trying inverse on non-square matrix
• Confusing error type
• Assuming code runs without error