Properties of Multiplication of Matrix

Last updated at Dec. 16, 2024 by

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

56.jpg

 

Commutativity in multiplication is not true

AB ≠ BA

 

Let’s solve them

 

AB

Properties of Multiplication of Matrix - Part 2

 

BA

Properties of Multiplication of Matrix - Part 3

 

Since

Properties of Multiplication of Matrix - Part 4

∴ AB ≠ BA

 

Zero matrix multiplication

 

Properties of Multiplication of Matrix - Part 5

 

We saw that

Properties of Multiplication of Matrix - Part 6

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

Properties of Multiplication of Matrix - Part 7

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

  (AB) C = O × C

      = O

 

A (BC)

Properties of Multiplication of Matrix - Part 8 .

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

 

A (B + C) = AB + AC

 

Properties of Multiplication of Matrix - Part 9

 

AB + AC

Properties of Multiplication of Matrix - Part 10

Therefore,

  A (B + C) = AB + AC

Let’s prove the next one

 

(A + B) C = AC + BC

Properties of Multiplication of Matrix - Part 11

 

 

Properties of Multiplication of Matrix - Part 12

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

Properties of Multiplication of Matrix - Part 13

 

Therefore,

  AI = IA = A

Properties of Multiplication of Matrix - Part 14

 

 

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Transcript

Let A = [■8(0&−1@0&2)] B = [■8(3&5@0&0)] C = [■8(3&0@0&4)] AB [■8(0&−1@0&2)] [■8(3&5@0&0)] = [■8(0×3+(−1)×0&0×5+(−1)×0@0×3+2×0&0×5+2×0)] = [■8(0&0@0&0)] BA [■8(3&5@0&0)] [■8(0&−1@0&2)] = [■8(3×0+5×0&3×(−1)+5×2@0×0+0×0&0×(−1)+0×2)] = [■8(0&−3+10@0&0)] = [■8(0&−7@0&0)] Since [■8(0&0@0&0)]≠[■8(0&−7@0&0)] ∴ AB ≠ BA Zero matrix multiplication For A = [■8(0&−1@0&2)], B = [■8(3&5@0&0)] We saw that AB = [■8(0&−1@0&2)][■8(3&5@0&0)] = [■8(0&0@0&0)] = O (AB) C AB = [■8(0&−1@0&2)] [■8(3&5@0&0)] = [■8(0×3+(−1)×0&0×5+(−1)×0@0×3+2×0&0×5+2×0)] = [■8(0&0@0&0)] (AB) C = [■8(0&0@0&0)] [■8(3&0@0&4)] = [■8(0&0@0&0)] A (BC) BC = [■8(3&5@0&0)] [■8(3&0@0&4)] = [■8(3×3+5×0&3×0+5×4@0×3+0×0&0×0+0×4)] = [■8(9&20@0&0)] A (BC) = [■8(0&−1@0&2)][■8(9&20@0&0)] = [■8(0×9+(−1)×0&0×20+(−1)×0@0×9+2×0&0×20+2×0)] = [■8(0&0@0&0)] Therefore, (AB) C = A (BC) A (B + C) B + C = [■8(3&5@0&0)] [■8(3&0@0&4)] = [■8(3+3&5+0@0+0&0+4)] = [■8(9&5@0&4)] A (B + C) = [■8(0&−1@0&2)][■8(9&5@0&4)] = [■8(0×9+(−1)×0&0×5+(−1)×4@0×9+2×0&0×5+2×4)] = [■8(0&−4@0&8)] AB + AC AB = [■8(0&−1@0&2)][■8(3&5@0&0)] = [■8(0×3+(−1)×0&0×5+(−1)×0@0×3+2×0&0×5+2×0)] = [■8(0&0@0&0)] AC = [■8(0&−1@0&2)][■8(3&0@0&4)] = [■8(0×3+(−1)×0&0×0+(−1)×4@0×3+2×0&0×0+2+4)] = [■8(0&−4@0&8)] AB + AC = [■8(0&0@0&0)] + [■8(0&−4@0&8)] = [■8(0&−4@0&8)] Therefore, A (B + C) = AB + AC Let’s prove the next one (A + B) C A + B = [■8(0&−1@0&2)] + [■8(3&5@0&0)] = [■8(0+3&(−1)+5@0+0&2+0)] = [■8(3&4@0&2)] (A + B) C = [■8(3&4@0&2)][■8(3&0@0&4)] = [■8(3×3+4×0&3×0+4×4@0×3+2×0&0×0+2×4)] = [■8(9&16@0&8)] AC + BC AC = [■8(0&−1@0&2)] [■8(3&0@0&4)] = [■8(0×3+(−1)×0&0×0+(−1)×4@0×3+2×0&0×0+2×4)] = [■8(0&−4@0&8)] BC = [■8(3&5@0&0)][■8(3&0@0&4)] = [■8(3×3+5×0&3×0+5×4@0×3+0×0&0×0+0×4)] = [■8(9&20@0&0)] AC + BC = [■8(0&−4@0&8)]+[■8(9&20@0&0)] = [■8(0+9&(−4)+20@0+0&8+0)] = [■8(9&16@0&8)] Therefore, (A + B) C = AC + BC For A = [■8(0&−1@0&2)] I = [■8(1&0@0&1)] AI = [■8(0&−1@0&2)] [■8(1&0@0&1)] = [■8(0×1+(−1)×0&0×0+(−1)×1@0×1+2×0&0×0+2×1)] = [■8(0&−1@0&2)] = A IA = [■8(1&0@0&1)][■8(0&−1@0&2)] = [■8(1×0+0×0&0×(−1)+0×2@0×0+1×0&0×(−1)+1×2)] = [■8(0&−1@0&2)] = A Note: For a 3 × 3 matrix, I will be a 3 × 3 matrix Example – For A = [■8(9&5&2@1&8&5@3&1&6)], I = [■8(1&0&0@0&1&0@0&0&1)]

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo