Matrix rank and null space in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of matrix rank calculation

What is the output of the following MATLAB code?

MATLAB

A = [1 2 3; 4 5 6; 7 8 9];
r = rank(A);
disp(r);

A 0

B 3

C 1

D 2

💡 Hint

Think about the linear dependence of rows in matrix A.

The matrix A has linearly dependent rows, so its rank is 2, not 3.

❓ Predict Output

intermediate

Null space dimension of a matrix

What is the size of the null space of matrix B?

MATLAB

B = [1 0 0; 0 1 0; 0 0 0];
N = null(B);
disp(size(N, 2));

A 2

B 0

C 1

D 3

💡 Hint

The null space dimension equals the number of free variables.

Matrix B has rank 2, so nullity = 3 - 2 = 1.

🔧 Debug

advanced

Identify the error in null space calculation

What error will this MATLAB code produce?

MATLAB

C = [1 2; 3 4];
N = null(C, 'r');
disp(N);

A Error: Undefined function or variable 'null'

B Error: Too many input arguments for null function

C No error, outputs null space basis

D Error: Matrix dimensions must agree

💡 Hint

Check the documentation for the null function's accepted arguments.

The null function in MATLAB does not accept a second argument 'r'.

🧠 Conceptual

advanced

Rank-nullity theorem application

Given a 4x5 matrix D with rank 3, what is the dimension of its null space?

A 2

B 3

C 1

D 4

💡 Hint

Use the formula: nullity = number_of_columns - rank.

Nullity = 5 - 3 = 2, so the null space dimension is 2.

❓ Predict Output

expert

Output of null space basis vectors

What is the output of this MATLAB code?

MATLAB

E = [1 2 3; 2 4 6; 3 6 9];
N = null(E);
disp(N);

A [ -0.8018; 0.2673; 0.5345 ]

B [ 0.2673; -0.8018; 0.5345 ]

C [ 0.5345; -0.8018; 0.2673 ]

D [ 0.5345; 0.2673; -0.8018 ]

💡 Hint

The null space basis vector is orthogonal to the rows of E.

The null function returns a basis vector approximately [-0.8018; 0.2673; 0.5345].