10 Math Problems: Multiplication of Matrices

Multiplication of Matrices

Let A and B be two matrices. Then, the multiplication of matrices A and B i.e. product of A and B is denoted by AB if it is defined. The product AB can be defined or cannot be defined. It depends on the order of A and B.

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Two matrices A and B are said to be conformable or compatible for the product AB if and only if the number of columns in A is equal to the number of rows in B. It means we can define the product AB only when the number of columns in A and number of rows in B are same. If this condition is not satisfied then the product of matrices A and B cannot be performed. In such a situation, matrices A and B are not conformable for multiplication.

If A and B are conformable for the product AB, then the number of rows in A followed by the number of columns in B gives the order of the product AB.

Let the order of A be m × n and that of B be n × p. Here, number of columns of A and the number of rows of B are the same. So, the matrices A and B are conformable for the product AB. Number of rows in A is m and number of columns in B is p. So, the order of product AB is m × p.

Order of the product of matrices A and B i.e. order of AB

If two matrices A and B are conformable or compatible for the product AB, it is not necessary that they are also conformable or compatible for the product BA. If A and B are square matrices of the same order, then they are conformable for the product AB as well as BA.

The element in row i and column j of the product AB is obtained by multiplying the elements in the i^th row of A by the corresponding elements in j^th column of B, and then by adding up the resulting products.

∴ (i, j)^th element of the product AB = Sum of the products of the elements of i^th row of A with the corresponding elements of j^th column of B.

Let A = (■(a_11&a_12@a_21&a_22 )) and B = (■(b_11&b_12@b_21&b_22 ))

Example: Order of the product of matrices A and B i.e. order of AB

Here, order of the product AB will be 2 × 2. If c_ij denotes the elements of the product matrix AB, then

AB = (■(c_11&c_12@c_21&c_22 )) Where, c11 = 1st row of A × 1st column of B = a11b11 + a12b21 c12 = 1st row of A × 2nd column of B = a11b12 + a12b22 c21 = 2nd row of A × 1st column of B = a21b11 + a22b21 c22 = 2nd row of A × 2nd column of B = a21b12 + a22b22 ∴ AB = (■(a_11 b_11+a_12 b_21&a_11 b_12+a_12 b_22@a_21 b_11+a_22 b_21&a_21 b_12+a_22 b_22 ))

Working Rule (Steps for multiplying matrices):

First of all guess the order of product AB and write down as follows:

AB = (■(a_11&a_12@a_21&a_22 ))(■(b_11&b_12@b_21&b_22 )) = (■( ……&…… @ ……&…… )) Then follow the steps given below: Step 1: (■(a_11&a_12@a_21&a_22 ))(■(b_11&b_12@b_21&b_22 )) = (■(a_11 b_11+a_12 b_21&……@……&……)) Step 2: (■(a_11&a_12@a_21&a_22 ))(■(b_11&b_12@b_21&b_22 )) = (■(a_11 b_11+a_12 b_21&a_11 b_12+a_12 b_22@……&……)) Step 3: (■(a_11&a_12@a_21&a_22 ))(■(b_11&b_12@b_21&b_22 )) = (■(a_11 b_11+a_12 b_21&a_11 b_12+a_12 b_22@a_21 b_11+a_22 b_21&……)) Step 4: (■(a_11&a_12@a_21&a_22 ))(■(b_11&b_12@b_21&b_22 )) = (■(a_11 b_11+a_12 b_21&a_11 b_12+a_12 b_22@a_21 b_11+a_22 b_21&a_21 b_12+a_22 b_22 ))

Examples:

1. Let A = (■(3&5@-1&2)) and B = (■(4@7)) be two matrices. Since A is of order 2 × 2 and B is of order 2 × 1, AB is defined and it is of order 2 × 1. AB = (■(3&5@-1&2))(■(4@7)) = (■(3.4+5.7@-1.4+2.7)) = (■(47@10)) Note that the matrix B is of order 2 × 1 and matrix A is of order 2 × 2, therefore, BA is not defined. 2. Let A = (■(1&5@2&0)) and B = (■(3&1@4&6)) be two matrices. Since A is of order 2 × 2 and B is of order 2 × 2, AB is defined and it is of order 2 × 2. AB = (■(1&5@2&0))(■(3&1@4&6)) = (■(1.3+5.4&1.1+5.6@2.3+0.4&2.1+0.6)) = (■(23&31@6&2)) BA = (■(3&1@4&6))(■(1&5@2&0)) = (■(3.1+1.2&3.5+1.0@4.1+6.2&4.5+6.0)) = (■(5&15@16&20)) Observe that AB ≠ BA.

Properties of Matrix Multiplication

1. Multiplication of matrices, generally, is not commutative, i.e. AB ≠ BA, in general.

i. When the matrix AB is defined, it not always necessary that BA can also be defined. For example, if the matrix A is m × n and the matrix B is n × p, AB exists whereas BA does not exist because p ≠ m.

ii. When both AB and BA are defined, it is not necessary that they should be of the same type. For example, if the matrix A is of order m × n and the matrix B is of order n × m, both the matrices AB and BA exist but AB is of order m × m while BA is of order n × n.

iii. When A and B are the square matrices of same order, then both AB and BA exist, but they are not necessarily equal. For example,

Let, A = (■(1&2@3&4)) and B = (■(1&2@4&3)). Then, AB = (■(1&2@3&4))(■(1&2@4&3)) = (■(1+8&2+6@3+16&6+12)) = (■(9&8@19&18)) BA = (■(1&2@4&3))(■(1&2@3&4)) = (■(1+6&2+8@4+9&8+12)) = (■(7&10@13&20)) Thus, AB ≠ BA.

iv. But, sometimes AB and BA are also equal. For example,

Let, A = (■(2&3@1&0)) and B = (■(0&-3@-1&2)). Then, AB = (■(2&3@1&0))(■(0&-3@-1&2)) = (■(0-3&-6+6@0+0&-3+0)) = (■(-3&0@0&-3)) BA = (■(0&-3@-1&2))(■(2&3@1&0)) = (■(0-3&0+0@-2+2&-3+0)) = (■(-3&0@0&-3)) Thus, AB = BA.

Hence in general AB ≠ BA.

2. Multiplication of matrices is associative i.e. if matrices A, B and C are conformable for multiplication, then (AB)C = A(BC).

3. Multiplication of matrices is distributive with respect to addition, i.e if matrices A, B and C are compatible for the requisite addition and multiplication, then A(B + C) = AB + AC and (A + B)C = AC + BC.

Let A = (■(1&1@2&1@1&2)), B = (■(0&-1@1&2)) and C = (■(1&0@2&1)). Then, B + C = (■(0&-1@1&2)) + (■(1&0@2&1)) = (■(1&-1@3&3)) A(B + C) = (■(1&1@2&1@1&2))(■(1&-1@3&3)) = (■(1+3&-1+3@2+3&-2+3@1+6&-1+6)) = (■(4&2@5&1@7&5)) ………….(i) AB = (■(1&1@2&1@1&2))(■(0&-1@1&2)) = (■(0+1&-1+2@0+1&-2+2@0+2&-1+4)) = (■(1&1@1&0@2&3)) AC = (■(1&1@2&1@1&2))(■(1&0@2&1)) = (■(1+2&0+1@2+2&0+1@1+4&0+2)) = (■(3&1@4&1@5&2)) AB + AC = (■(1&1@1&0@2&3)) + (■(3&1@4&1@5&2)) = (■(4&2@5&1@7&5)) …………………...(ii) From (i) and (ii), we get, A(B + C) = AB + AC. Similarly, we can verify that, (A + B)C = AC + BC.

4. If A is a square matrix and I is an identity matrix of same order, then AI = IA = A.

Let A = (■(1&2@3&4)) and I = (■(1&0@0&1)). Then, AI = (■(1&2@3&4))(■(1&0@0&1)) = (■(1+0&0+2@3+0&0+4)) = (■(1&2@3&4)) IA = (■(1&0@0&1))(■(1&2@3&4)) = (■(1+0&2+0@0+3&0+4)) = (■(1&2@3&4)) ∴ AI = IA = A. Here, matrix I is called multiplicative identity.

Note: Multiplication of matrices gives us some results which are different from the results obtained in the case of numbers. Some of these results illustrated in the following examples:

i. If AB is a null matrix, it does not imply that at least one of the matrices A or B must be a zero matrix. For example,