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Chapter 3 · Matrices

Exercise 3.1 · Q10 (MCQ): How many 3×33\times3 matrices have every entry 0 or 1?

Exercise 3.1Order of a matrix · Section 3.2
The question

The number of all possible matrices of order 3×33\times3 with each entry 0 or 1 is:

(A) 27
(B) 18
(C) 81
(D) 512

Exam answer

A 3×33\times3 matrix has 9 entries, and each entry can be chosen in 2 ways (0 or 1).

∴\therefore Number of matrices =29=512= 2^9 = 512⁠.

Answer: (D) 512

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A 3×33\times3 matrix has 99 entries. Building one means making 9 separate choices, one per entry, and each choice has 2 options (0 or 1). Choices made one after another, independently, multiply: there are 2×2×⋯×2=292\times2\times\cdots\times2 = 2^9 ways in all. And two matrices are different as soon as one entry differs, so different choices give different matrices.

Given, to find, and the plan
Given
Matrices of order 3×33\times3⁠; each entry must be 0 or 1.
To find
How many different such matrices exist.
Plan
(1) Count the entries. (2) Use equality of matrices to see that a matrix is exactly a list of 9 choices. (3) Count the lists with the multiplication principle. (4) Test the method on small cases and examine every option.

Basics you need

1Number of entries

An m×nm\times n matrix has mnmn entries, so a 3×33\times3 matrix has 9. Basics: rows, columns and the address

2When two matrices are the same

Two matrices of the same order are equal exactly when every pair of corresponding entries is equal (Definition 2). So changing even one entry gives a different matrix. Lesson: equality of matrices

3Multiplication principle

If one choice can be made in pp ways and, whatever it was, a second choice can be made in qq ways, then the pair of choices can be made in p×qp\times q ways. With 9 choices of 2 ways each: 292^9 ways.

4Powers of 2

21=22^1 = 2⁠, 22=42^2 = 4⁠, 23=82^3 = 8⁠, 24=162^4 = 16⁠, …, 29=5122^9 = 512⁠.

Solution

  1. Count the entries

    A 3×33\times3 matrix has 3 rows and 3 columns, so 3×3=93\times3 = 9 entries:

    [a11a12a13a21a22a23a31a32a33]\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}
    The question restricts the entries, not the order, so the shape is fixed and only the 9 entries vary.
    An m×nm\times n matrix has mnmn entries (Section 3.2.1). Basics: rows and columns

  2. A matrix is the same thing as a list of 9 choices

    Each matrix of this kind is fixed completely by choosing a value, 0 or 1, for each of a11,a12,…,a33a_{11}, a_{12}, \dots, a_{33}⁠. Conversely, by Definition 2, two such matrices are equal only if all 9 entries agree. So:

    • every list of 9 choices gives a matrix, and
    • different lists give different matrices.

    Hence counting the matrices is the same as counting the lists of 9 choices.

    This step makes sure we are neither double-counting nor missing anything. Without it, one might worry that two different choices could produce "the same" matrix; Definition 2 rules that out, because the matrices would differ in at least one entry.
    Equal matrices agree entry by entry. Lesson: equality of matrices

  3. Count the choices

    Choose the entries one at a time. Each has 2 options, whatever was chosen for the others:

    2×2×2×2×2×2×2×2×2⏟9 entries=29.\underbrace{2\times2\times2\times2\times2\times2\times2\times2\times2}_{9\ \text{entries}} = 2^9.

    Working out the power: 23=82^3 = 8⁠, so

    29=(23)3=83=512.2^9 = (2^3)^3 = 8^3 = 512.

    So there are 512 such matrices.

    1 0 / 1 2 0 / 1 3 0 / 1 4 0 / 1 5 0 / 1 6 0 / 1 7 0 / 1 8 0 / 1 9 0 / 1 9 entries each one: 0 or 1 (2 choices) choices are independent, so 2 × 2 × … × 2 (nine 2s) = 2⁹ = 512 matrices
    Nine cells, each filled independently with 0 or 1. Every cell doubles the number of possibilities, so nine cells give 29=5122^9 = 512 different matrices.
    The choices are independent: picking a11=1a_{11} = 1 does not restrict a12a_{12} in any way. That is exactly when the multiplication principle applies, and the counts multiply rather than add.
    Multiplication principle: independent choices multiply.

  4. Test the method on small cases

    A 1×11\times1 matrix with entry 0 or 1: [0][0]⁠, [1][1]⁠, so 2=212 = 2^1⁠. A 1×21\times2 matrix: four possibilities, 0 0, 0 1, 1 0, 1 1, so 4=224 = 2^2⁠. In general an m×nm\times n matrix with entries 0 or 1 can be filled in 2mn2^{mn} ways. A 2×22\times2 one gives 24=162^4 = 16⁠, and our 3×33\times3 gives 29=5122^9 = 512⁠.

    Small cases can be listed by hand, and they match the formula, which confirms that the reasoning (one factor of 2 per entry) is right.

  5. Examine every option

    OptionNumberVerdict
    (A)27Wrong. 27=3327 = 3^3⁠, as if 3 choices were made for 3 things. But there are 9 entries with 2 choices each, not 3 with 3.
    (B)18Wrong. 18=9×218 = 9\times2⁠, from multiplying the number of entries by the number of values. That counts "which entry and which value" for a single entry; it does not count whole matrices.
    (C)81Wrong. 81=92=3481 = 9^2 = 3^4⁠, which puts the 9 and the 2 the wrong way round: 929^2 instead of 292^9⁠.
    (D)512Correct. 29=5122^9 = 512⁠, from Step 3.
    The base of the power is the number of options per entry (2), and the exponent is the number of entries (9). Each wrong option comes from mixing up these roles, or from adding or multiplying the wrong numbers.
Result

The correct answer is (D): there are 29=5122^9 = 512 matrices of order 3×33\times3 with each entry 0 or 1. (Textbook answer: D.)

Check it with numbers

Think of each matrix as a 9-digit binary string, reading the entries row by row (for example 100010001 for the identity matrix). Binary strings of length 9 represent exactly the whole numbers from 0 to 29−1=5112^9 - 1 = 511⁠, one each, so there are 512512 of them ✓. The site's verification script also lists all such matrices by computer and counts 512 ✓.

Common mistakes
  1. Writing 929^2 instead of 292^9⁠. The exponent counts the choices; the base counts the options per choice.
  2. Adding instead of multiplying, which gives 2+2+⋯+2=182 + 2 + \cdots + 2 = 18⁠.
  3. Thinking the order of the entries does not matter. A 1 in position a12a_{12} and a 1 in position a21a_{21} give different matrices.
What to take away
  1. An m×nm\times n matrix whose entries each come from a set of kk values can be filled in kmnk^{mn} ways.
  2. Different entries anywhere give different matrices, by the definition of equality.
  3. Independent choices multiply.

Every fact used, and where it comes from

FactUsed inSource
An m×nm\times n matrix has mnmn elementsStep 1Section 3.2.1 (Lesson: what a matrix is)
Two matrices are equal only if all corresponding entries are equalStep 2Definition 2, Section 3.3.1 (Lesson: equality of matrices)
Multiplication principle of countingStep 3NCERT Class XI Mathematics, Chapter 6 (Permutations and Combinations), Section 6.2, Fundamental Principle of Counting
29=5122^9 = 512⁠; the count 512 by listing all matricesSteps 3, 5; CheckRecomputed in the site's verification script
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