Matrix not only has addition and subtraction but also it has multiplication. There are some rules of multiplication of matrix depend on the problem. Matrix has some type of multiplication problem:

Table of Contents

**Multiplication with a constant (Scalar Multiplication)**

Multiplication of matrix with a constant has simple concept. It is just multiplying each element of matrix with the constant. If there is any real number as a constant (it is also called scalar) and a matrix with any dimension,

Scalar = 3 and matrix A

**Matrices Multiplication **

Matrices multiplication means multiplying matrix with another matrix. It can not apply in any matrix. There are some rules:

- It is only matrices that has certain dimension that can be operated. It means the dimension of the first matrix is
*pxq*then the second matrix must be*qxr*. The number of columns in first matrix must be same with number of rows in second matrix. Then the result (matrix product) has dimension*pxr*. Number of rows is like first matrix and number of columns is like second matrix.

If matrix A_{1×3}and matrix B_{3×2}then the result of A.B is C_{1×2}. - If there are matrix A and B then AxB ≠ BxA, in another word AB ≠ BA.
- Matrices multiplication has own rule. It is different with algebra operation and scalar multiplication. The base concept is dot product.

Concept of dot product:

(1,3,5)·(6,7,8) = 1.6 + 3.7 + 5.8 = 6 + 21 + 40 = 67

If there are matrix A_{2×2} and B_{2×2}, then the result of AxB can be solved by using steps:

a. Multiply first row of matrix A with the first column of matrix B. the result will be element *a _{1,1}*.

b. Multiply first row of matrix A with the second column of matrix B. the result will be element of

*a*.

_{1,2}c. Multiply second row of matrix A with the first column of matrix B. the result will be element of

*a*.

_{2,1}d. Do the same way to get other element of the result.

Remember that the dimension of the result came from matrix A_{2}_{x}_{2} and matrix B_{2}_{x}_{2} then the result is matrix C_{2×2} (the red one). The blue one shows the requirement (they must be same) that two matrices can be multiply.

**Identity Matrix**

Matrix has an identity. It is symbolized as “I”.

For matrix that has larger dimension is not difficult. The concept is “1” is always in the diagonal. The others are “0”. Result of multiplying matrix with identity matrix is the matrix itself.

If there is matrix A, then

**A x I = A**

**I x A = A**

**Matrices Multiplication Example**

- Determine the result of multiplying of A.B if

2. Determine the result if M.I + 3N – M if

Because of multiplying between M and identity get matrix M itself, then

M.I + 3N – M = M + 3N – M = 3N

The answer is matrix 3N.

3. Determine the result of L.K and the dimension if

Dimension of matrix L is 2×3

Dimension of matrix K is 3×1

Then the result has dimension 2×1

4. There are any matrix 3×2 and 2×1.

Is it possible to multiply?

(If you answer “yes” then give an example).

It is possible.

The dimension of the first matrix is *3×2* then the second matrix is *2×1*. The number of columns in first matrix is same with number of rows in second matrix. Then the result (matrix product) has dimension *3×1*.

Example (it can be different with others)

5. Determine the result of multiplication of matrix below.

6. If there are some matrix and XY = Z, determine the value of 2k.

XY = Z

Then

So, k = -11 and 2k = 2(-11) = -22

7. There are matrixes

Determine p + q + r + s.

Because of

Then

- 3r = 3 → r = 1
- -5r = -3 – q

-5(1) = -3 – q

-5 = -3 – q

q = -3 + 5 = 2

- 3p – 1 = q + 3

3p = 2 + 3 + 1

3p = 6

p = 2

- 1 – p = -5 + s

1 – 2 = -5 + s

-1 + 5 = s

s = 4

so, p + q + r + s = 2 + 2 + 1 + 4 = 9

8. Determine the result of matrix multiplication

(AB^{-1}) (B + A) A^{-1} = …

(AB^{-1}) (B + A) A^{-1} = (AB^{-1}) (BA^{-1} + AA^{-1})

= (AB^{-1}) (BA^{-1} + I)

= AB^{-1}BA^{-1} + AB^{-1}

= AA^{-1} + AB^{-1}

= I + AB^{-1}

9. There is a multiplication of matrixes

If C is a matrix 2×2 and the determinant is 10, determine a + 7b.

Determinant of C = ad – bc

10 = (-2)(a+2b) – (-2)(-5b+2)

10 = (-2a-4b) – (10b – 4)

10 = -2a – 4b – 10b + 4

10 = -2a -14b + 4

10 – 4 = -2a – 14b

6 = -2a – 14b (divide by 2)

3 = -a – 7b

3 = -(a+7b)

-3 = a + 7b