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NCERT Exemplar · Q81

Q.Matrix addition is associative as well as commutative.

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Matrix addition is both associative and commutative because it is defined element-wise, and the addition of real numbers (or scalars) is itself associative and commutative. So for any matrices AA, BB, CC of the same order, we have (A+B)+C=A+(B+C)(A+B)+C = A+(B+C) and A+B=B+AA+B = B+A.

Why This Works: The Core Idea

Matrix addition is the simplest operation in linear algebra — and that simplicity is exactly why these properties hold. When you add two matrices, you add their corresponding entries one by one. There is no cross-talk between entries, no multiplication, no fancy rules. Each entry behaves independently.

So the question of whether matrix addition is associative or commutative reduces to a much simpler question: Is the addition of ordinary numbers associative and commutative? The answer is yes — and that's the entire proof.

Let's see this in detail.

Step-by-Step Reasoning

1. Define what we mean by matrix addition.

If AA and BB are both m×nm \times n matrices, their sum A+BA+B is another m×nm \times n matrix where each entry is the sum of the corresponding entries:

(A+B)ij=Aij+Bij(A+B)_{ij} = A_{ij} + B_{ij}

for every row ii and column jj. This definition is the foundation of everything that follows.

2. Check commutativity: A+B=B+AA+B = B+A.

Take any entry at position (i,j)(i,j) in the sum A+BA+B. By definition, it is Aij+BijA_{ij} + B_{ij}. Now consider the same entry in B+AB+A: it is Bij+AijB_{ij} + A_{ij}.

Since addition of real numbers is commutative, Aij+Bij=Bij+AijA_{ij} + B_{ij} = B_{ij} + A_{ij} for every ii and jj. Therefore every entry of A+BA+B equals the corresponding entry of B+AB+A, so the matrices are identical:

A+B=B+AA+B = B+A

Tip

This is the cleanest proof in linear algebra — you never need to write out entire matrices. Just point to a single arbitrary entry and use the commutativity of real numbers.

3. Check associativity: (A+B)+C=A+(B+C)(A+B)+C = A+(B+C).

Now let AA, BB, CC all be m×nm \times n matrices. Consider the entry at position (i,j)(i,j) in (A+B)+C(A+B)+C.

First, (A+B)ij=Aij+Bij(A+B)_{ij} = A_{ij} + B_{ij}. Then adding CC gives:

((A+B)+C)ij=(Aij+Bij)+Cij((A+B)+C)_{ij} = (A_{ij} + B_{ij}) + C_{ij}

Now look at the same entry in A+(B+C)A+(B+C). First, (B+C)ij=Bij+Cij(B+C)_{ij} = B_{ij} + C_{ij}. Then adding AA gives:

(A+(B+C))ij=Aij+(Bij+Cij)(A+(B+C))_{ij} = A_{ij} + (B_{ij} + C_{ij})

Since addition of real numbers is associative, (Aij+Bij)+Cij=Aij+(Bij+Cij)(A_{ij} + B_{ij}) + C_{ij} = A_{ij} + (B_{ij} + C_{ij}) for every ii and jj. Hence every entry matches, and the matrices are equal:

(A+B)+C=A+(B+C)(A+B)+C = A+(B+C)

Watch out

A common mistake is to think that because matrix multiplication is not commutative, matrix addition might also fail to be commutative. That's wrong — addition and multiplication are completely different operations. Always check the definition before assuming anything.

4. What about the zero matrix?

The m×nm \times n zero matrix, denoted 00, has all entries equal to 00. It acts as the additive identity: …

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