Multiplication of matrices

Chapter 3 Class 12 Matrices
Concept wise

Let’s look at some properties of multiplication of matrices.

1. Commutativity is not true:

AB ≠ BA

2. Zero matrix on multiplication

If AB = O,

then A ≠ O, B ≠ O is possible

3. Associative law:

(AB) C = A (BC)

4. Distributive law:

A (B + C) = AB + AC

(A + B) C = AC + BC

•

5. Multiplicative identity:

For a square matrix A

AI = IA = A

where I is the identity matrix of the same order as A.

Let’s look at them in detail

We used these matrices

AB ≠ BA

Let’s solve them

AB

BA

Since

∴ AB ≠ BA

## Zero matrix multiplication

We saw that

So, AB = O

But A ≠ O & B ≠ O

Therefore,

If two matrices multiply to become zero matrix,

then it is not true that A = O or B = O

Note: This is different from numbers

If ab = 0,
then either a = 0 or b = 0

But this is not true for matrices

## Associative law

(AB) C = A (BC)

Let’s solve this

(AB) C

Note: Any matrix multiplied to zero matrix is a zero matrix

(AB) C = O × C

= O

A (BC)

.

Therefore,

(AB) C = A (BC)

## Distributive law

Distributive law says that -

• A (B + C) = AB + AC
• (A + B) C = AC + BC

Let’s prove both of them

### AB + AC

Therefore,

A (B + C) = AB + AC

Let’s prove the next one

### (A + B) C = AC + BC

Therefore,

(A + B) C = AC + BC

## Multiplicative Identity

For any square matrix A,

AI = IA = A

Where I is identity matrix of same order as A

Therefore,

AI = IA = A

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### Transcript

Let A = [■8(0&−[email protected]&2)] B = [■8(3&[email protected]&0)] C = [■8(3&[email protected]&4)] AB [■8(0&−[email protected]&2)] [■8(3&[email protected]&0)] = [■8(0×3+(−1)×0&0×5+(−1)×[email protected]×3+2×0&0×5+2×0)] = [■8(0&[email protected]&0)] BA [■8(3&[email protected]&0)] [■8(0&−[email protected]&2)] = [■8(3×0+5×0&3×(−1)+5×[email protected]×0+0×0&0×(−1)+0×2)] = [■8(0&−[email protected]&0)] = [■8(0&−[email protected]&0)] Since [■8(0&[email protected]&0)]≠[■8(0&−[email protected]&0)] ∴ AB ≠ BA Zero matrix multiplication For A = [■8(0&−[email protected]&2)], B = [■8(3&[email protected]&0)] We saw that AB = [■8(0&−[email protected]&2)][■8(3&[email protected]&0)] = [■8(0&[email protected]&0)] = O (AB) C AB = [■8(0&−[email protected]&2)] [■8(3&[email protected]&0)] = [■8(0×3+(−1)×0&0×5+(−1)×[email protected]×3+2×0&0×5+2×0)] = [■8(0&[email protected]&0)] (AB) C = [■8(0&[email protected]&0)] [■8(3&[email protected]&4)] = [■8(0&[email protected]&0)] A (BC) BC = [■8(3&[email protected]&0)] [■8(3&[email protected]&4)] = [■8(3×3+5×0&3×0+5×[email protected]×3+0×0&0×0+0×4)] = [■8(9&[email protected]&0)] A (BC) = [■8(0&−[email protected]&2)][■8(9&[email protected]&0)] = [■8(0×9+(−1)×0&0×20+(−1)×[email protected]×9+2×0&0×20+2×0)] = [■8(0&[email protected]&0)] Therefore, (AB) C = A (BC) A (B + C) B + C = [■8(3&[email protected]&0)] [■8(3&[email protected]&4)] = [■8(3+3&[email protected]+0&0+4)] = [■8(9&[email protected]&4)] A (B + C) = [■8(0&−[email protected]&2)][■8(9&[email protected]&4)] = [■8(0×9+(−1)×0&0×5+(−1)×[email protected]×9+2×0&0×5+2×4)] = [■8(0&−[email protected]&8)] AB + AC AB = [■8(0&−[email protected]&2)][■8(3&[email protected]&0)] = [■8(0×3+(−1)×0&0×5+(−1)×[email protected]×3+2×0&0×5+2×0)] = [■8(0&[email protected]&0)] AC = [■8(0&−[email protected]&2)][■8(3&[email protected]&4)] = [■8(0×3+(−1)×0&0×0+(−1)×[email protected]×3+2×0&0×0+2+4)] = [■8(0&−[email protected]&8)] AB + AC = [■8(0&[email protected]&0)] + [■8(0&−[email protected]&8)] = [■8(0&−[email protected]&8)] Therefore, A (B + C) = AB + AC Let’s prove the next one (A + B) C A + B = [■8(0&−[email protected]&2)] + [■8(3&[email protected]&0)] = [■8(0+3&(−1)[email protected]+0&2+0)] = [■8(3&[email protected]&2)] (A + B) C = [■8(3&[email protected]&2)][■8(3&[email protected]&4)] = [■8(3×3+4×0&3×0+4×[email protected]×3+2×0&0×0+2×4)] = [■8(9&[email protected]&8)] AC + BC AC = [■8(0&−[email protected]&2)] [■8(3&[email protected]&4)] = [■8(0×3+(−1)×0&0×0+(−1)×[email protected]×3+2×0&0×0+2×4)] = [■8(0&−[email protected]&8)] BC = [■8(3&[email protected]&0)][■8(3&[email protected]&4)] = [■8(3×3+5×0&3×0+5×[email protected]×3+0×0&0×0+0×4)] = [■8(9&[email protected]&0)] AC + BC = [■8(0&−[email protected]&8)]+[■8(9&[email protected]&0)] = [■8(0+9&(−4)[email protected]+0&8+0)] = [■8(9&[email protected]&8)] Therefore, (A + B) C = AC + BC For A = [■8(0&−[email protected]&2)] I = [■8(1&[email protected]&1)] AI = [■8(0&−[email protected]&2)] [■8(1&[email protected]&1)] = [■8(0×1+(−1)×0&0×0+(−1)×[email protected]×1+2×0&0×0+2×1)] = [■8(0&−[email protected]&2)] = A IA = [■8(1&[email protected]&1)][■8(0&−[email protected]&2)] = [■8(1×0+0×0&0×(−1)+0×[email protected]×0+1×0&0×(−1)+1×2)] = [■8(0&−[email protected]&2)] = A Note: For a 3 × 3 matrix, I will be a 3 × 3 matrix Example – For A = [■8(9&5&[email protected]&8&[email protected]&1&6)], I = [■8(1&0&[email protected]&1&[email protected]&0&1)]