Why numerical computation solves real problems in MATLAB - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of matrix inversion with numerical precision

What is the output of this MATLAB code snippet?

MATLAB

A = [1 2; 3 4];
B = inv(A);
disp(B * A);

A [1.0000 0.0000; 0.0000 1.0000]

B [1 0; 0 1]

C [1.0000 0.0000; 0.0000 0.9999]

D [0 0; 0 0]

💡 Hint

Matrix multiplied by its inverse should be identity matrix.

Multiplying a matrix by its inverse results in the identity matrix. MATLAB displays floating point numbers with 4 decimals by default.

🧠 Conceptual

intermediate

Why numerical methods approximate solutions

Why do numerical computations often provide approximate rather than exact solutions in real problems?

A Because numerical computations ignore rounding errors by design.

B Because numerical methods always use symbolic algebra for exact answers.

C Because computers cannot represent irrational numbers exactly and use finite precision.

D Because numerical methods only work for integer inputs.

💡 Hint

Think about how computers store numbers.

Computers store numbers with limited digits, so irrational or very precise numbers are approximated, causing numerical methods to produce approximate results.

🔧 Debug

advanced

Identify the error in numerical integration code

What error does this MATLAB code produce when run?

MATLAB

f = @(x) x.^2;
result = integral(f, 0, 1, 'AbsTol', -1);
disp(result);

A Error: AbsTol must be positive

B Error: Missing semicolon

C Output: 0.3333

D Error: Function handle is invalid

💡 Hint

Check the tolerance parameter value.

The 'AbsTol' parameter for integral must be a positive number. Negative values cause an error.

📝 Syntax

advanced

Correct syntax for element-wise operations

Which option correctly computes the element-wise square of vector v in MATLAB?

MATLAB

v = [1, 2, 3, 4];

A v**2

B v^2

C square(v)

D v.^2

💡 Hint

Remember MATLAB uses dot for element-wise operations.

In MATLAB, the dot operator before ^ means element-wise power. Without dot, ^ is matrix power which is invalid for vectors.

🚀 Application

expert

Result of iterative numerical method for root finding

What is the approximate value of x after 3 iterations of Newton's method starting at x=1 for f(x)=x^2-2?

MATLAB

f = @(x) x^2 - 2;
f_prime = @(x) 2*x;
x = 1;
for i = 1:3
    x = x - f(x)/f_prime(x);
end
disp(x);

A 2.0

B 1.4142

C 1.5

D 1.25

💡 Hint

Newton's method converges quickly to sqrt(2).

Starting at 1, Newton's method for f(x)=x^2-2 converges to sqrt(2) ≈ 1.4142 after 3 iterations.