Colon operator for ranges in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a vector from 1 to 5 using the colon operator.

MATLAB

v = 1[1]5;

Drag options to blanks, or click blank then click option'

A,

B;

C.

D:

Attempts:

3 left

💡 Hint

Common Mistakes

Using a semicolon instead of a colon.

Using a comma which separates elements but does not create a range.

The colon operator : creates a range from the start value to the end value.

2fill in blank

medium

Complete the code to create a vector from 2 to 10 with step size 2.

MATLAB

v = 2[1]2[2]10;

Drag options to blanks, or click blank then click option'

A:

B;

C,

D.

Attempts:

3 left

💡 Hint

Common Mistakes

Using commas instead of colons.

Using only one colon which assumes step size 1.

The colon operator with three parts start:step:end creates a vector starting at start, increasing by step, up to end.

3fill in blank

hard

Fix the error in the code to create a vector from 5 down to 1 using the colon operator.

MATLAB

v = 5[1]1;

Drag options to blanks, or click blank then click option'

A-:

B:

C:-

D;

Attempts:

3 left

💡 Hint

Common Mistakes

Using invalid operators like '-:' or ':-'.

Using semicolon which ends a statement.

The colon operator can create decreasing sequences if the start is greater than the end and the step is negative. Here, the default step is +1, so to count down, you must specify the step explicitly.

However, the question only asks to fix the operator, so the colon is correct. To count down, the full syntax is 5:-1:1.

4fill in blank

hard

Complete the code to create a vector from 10 down to 2 with step size 2 using the colon operator.

MATLAB

v = 10:[1]2:2;

Drag options to blanks, or click blank then click option'

A:

B-

C-1

D;

Attempts:

3 left

💡 Hint

Common Mistakes

Using semicolon instead of colon.

Using positive step when counting down.

To count down by 2, the step is negative 2, written as -2. Here, the step is constructed by combining the minus sign and the number 2.

5fill in blank

hard

Fill both blanks to create a vector from 1 to 9 with step size 3 using the colon operator.

MATLAB

v = [1]:[2]9;

Drag options to blanks, or click blank then click option'

A1

B:

C3

D;

Attempts:

3 left

💡 Hint

Common Mistakes

Using semicolon instead of colon.

Mixing up the order of start, step, and end.

Here, start is 1, step is 3, and end is 9.

Practice

1. What does the MATLAB expression 1:5 produce?

easy

A. [1 3 5]

B. [5 4 3 2 1]

C. [1 2 3 4 5]

D. [1 2 3 4]

Solution

1. Step 1: Understand the colon operator syntax

   The expression start:end creates a sequence from start to end with step 1 by default.

2. Step 2: Apply to given expression

   Here, 1:5 means numbers from 1 to 5 increasing by 1: 1, 2, 3, 4, 5.

3. Final Answer:

   [1 2 3 4 5] -> Option C

4. Quick Check:

   1 to 5 by 1 = [1 2 3 4 5] [OK]

Hint: Colon with two numbers counts up by 1 [OK]

Common Mistakes:

• Thinking it skips numbers
• Confusing start and end order
• Assuming it counts down automatically

2. Which of the following is the correct syntax to create a vector from 10 down to 2 with step -2 in MATLAB?

easy

A. 10:-2:2

B. 10:2:-2

C. 10--2:2

D. 10:-2:-2

Solution

1. Step 1: Recall colon operator with step syntax

   The syntax is start:step:end. For counting down, step must be negative.

2. Step 2: Check each option

   10:-2:2 uses 10:-2:2, which counts down from 10 to 2 by -2 correctly.

3. Final Answer:

   10:-2:2 -> Option A

4. Quick Check:

   Descending with negative step = 10:-2:2 [OK]

Hint: Use negative step for descending ranges [OK]

Common Mistakes:

• Putting step after end value
• Using positive step for descending
• Syntax errors with double minus signs

3. What is the output of the MATLAB code:

vec = 3:3:12;

disp(vec);

medium

A. [3 4 5 6 7 8 9 10 11 12]

B. [3 6 9 12]

C. [3 6 9]

D. [12 9 6 3]

Solution

1. Step 1: Understand the colon operator with step

   The syntax start:step:end creates numbers from start to end increasing by step.

2. Step 2: Calculate the sequence

   Starting at 3, add 3 each time: 3, 6, 9, 12. Since 12 is included, the sequence ends there.

3. Final Answer:

   [3 6 9 12] -> Option B

4. Quick Check:

   3 to 12 by 3 = [3 6 9 12] [OK]

Hint: Multiply step by count until reaching or passing end [OK]

Common Mistakes:

• Stopping before including end if exact
• Confusing step with end value
• Assuming step is always 1

4. Identify the error in this MATLAB code:

range = 5:0.5:3;

medium

A. Empty vector because start > end with positive step

B. Syntax error due to decimal step

C. Vector counts down correctly

D. Step must be integer

Solution

1. Step 1: Analyze start, step, and end values

   Start is 5, end is 3, step is 0.5 (positive).

2. Step 2: Understand colon operator behavior

   With positive step, MATLAB counts up. Since 5 > 3, no values satisfy the condition, so the vector is empty.

3. Final Answer:

   Empty vector because start > end with positive step -> Option A

4. Quick Check:

   Positive step but start > end = empty vector [OK]

Hint: Step sign must match direction from start to end [OK]

Common Mistakes:

• Assuming decimal steps cause syntax errors
• Expecting vector to count down with positive step
• Thinking step must be integer

5. You want to create a vector in MATLAB that starts at 0, goes up to 1, but only includes values less than 1 with step 0.2. Which expression correctly does this?

hard

A. 0:0.2:1

B. 0:0.2:<1

C. 0:0.2:1.1

D. 0:0.2:0.9

Solution

1. Step 1: Understand the requirement

   The vector should start at 0, increase by 0.2, and include values less than 1 (not including 1).

2. Step 2: Check which end value excludes 1

   Using 0:0.2:1 includes 1 since it matches the sequence (0, 0.2, 0.4, 0.6, 0.8, 1). Since 1 is included, it violates the condition.

3. Step 3: Choose end value less than 1

   Using 0:0.2:0.9 stops at 0.8 (0.8 + 0.2 = 1.0 > 0.9), which is less than 1, so it meets the condition.

4. Final Answer:

   0:0.2:0.9 -> Option D

5. Quick Check:

   End less than 1 excludes 1 from vector [OK]

Hint: Set end value just below the limit to exclude it [OK]

Common Mistakes:

• Using end value equal to limit includes it
• Trying invalid syntax like <1
• Assuming floating point precision excludes end automatically