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NCERT Exemplar · Q69

Q.Matrix multiplication is _________ over addition.

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Matrix multiplication is distributive over addition. This means for matrices AA, BB, CC of compatible sizes, we have A(B+C)=AB+ACA(B+C) = AB + AC and (B+C)A=BA+CA(B+C)A = BA + CA.

Why This Property Matters

Matrix multiplication and addition are the two fundamental operations you'll use constantly in linear algebra. The question asks about their relationship — specifically, what happens when you multiply a matrix by a sum of two others.

Think of it like arithmetic with numbers: 3×(4+5)=3×4+3×53 \times (4 + 5) = 3 \times 4 + 3 \times 5. That's distributivity. For matrices, the same idea holds, but with a crucial twist: order matters because matrix multiplication is not commutative. So we actually get two distributive laws — one for left-multiplication and one for right-multiplication.

Watch out

A common mistake is to assume A(B+C)=AB+ACA(B+C) = AB + AC works for any three matrices. It only works when the sizes are compatible: AA must be m×nm \times n, and both BB and CC must be n×pn \times p (so B+CB+C is defined, and AA can multiply it). Always check dimensions first.

Step-by-Step Verification

Let's prove the left distributive law: A(B+C)=AB+ACA(B+C) = AB + AC.

1. Set up the matrices.

Let AA be an m×nm \times n matrix, and BB and CC be n×pn \times p matrices. Then B+CB+C is also n×pn \times p, so A(B+C)A(B+C) is m×pm \times p — same as ABAB and ACAC, which are also m×pm \times p. The dimensions match, so the equality is at least possible.

2. Write the (i,j)(i,j) entry of A(B+C)A(B+C).

The (i,j)(i,j) entry of a product is the dot product of row ii of the first matrix with column jj of the second. So:

[A(B+C)]ij=∑k=1nAik(B+C)kj[A(B+C)]_{ij} = \sum_{k=1}^{n} A_{ik} (B+C)_{kj}

3. Use the definition of matrix addition.

The (k,j)(k,j) entry of B+CB+C is simply Bkj+CkjB_{kj} + C_{kj}. So:

[A(B+C)]ij=∑k=1nAik(Bkj+Ckj)[A(B+C)]_{ij} = \sum_{k=1}^{n} A_{ik} (B_{kj} + C_{kj})

4. Distribute the sum over the addition inside.

This is just algebra with real numbers (the entries are numbers):

[A(B+C)]ij=∑k=1n(AikBkj+AikCkj)[A(B+C)]_{ij} = \sum_{k=1}^{n} (A_{ik} B_{kj} + A_{ik} C_{kj})

5. Split the sum into two separate sums.

A sum of a sum is the sum of the individual sums: …

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