Multiplication and Power of Matrices

   

The multiplications of matrices are presented using examples and questions with solutions.

Multiplication of Rows and Columns Matrices

Let A be a row matrix of order 1 × p with entries a 1j and B be a column matrix of order p × 1 with entries b j1 . The multiplication of matrix A by matrix B is a 1 × 1 matrix defined by:

Example 1
Matrices A and B are defined by

Find the matrix A B.

Solution

Multiplication of Matrices

We now apply the idea of multiplying a row by a column to multiplying more general matrices. Let A be an m × p matrix and B be an p × n matrix. Let R 1 , R 2 , ... R m be the rows of matrix A and C 1 , C 2 , ... C n be the columns of column B and write the two matrices as:
$$A = \begin{bmatrix} R_1 \\ R2 \\ . \\ .\\ R_m \end{bmatrix}$$ and $$B = \begin{bmatrix} C_1 & C_2 &...& C_n \\ \end{bmatrix}$$
The product of the two matrices A and B is matrix C of order m × n defined by
$C = A \cdot B = \begin{bmatrix} R_1 . C_1 & R_1 . C_2 & ... &R_1 . C_n \\ R_2 . C_1 & R_2 . C_2 & ... & R_2 . C_n \\ && ... & \\ && ... &\\ && ... &\\ R_m . C_1 & R_m . C_2 & ... &R_m . C_n \end{bmatrix}$
and an entry $${ij}$$ of C is given by: $$c_{ij} = R_i \cdot C_j = \sum_{k=1}^{p} a_{ik} \times b_{kj}$$

Example 2
Find the product
$$\begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} 2 & - 2 \\ 1 & 0 \\ -1 & 2 \end{bmatrix}$$

Solution
The matrix on the left has 2 rows R
1 and R 2 the matrix on the right has 2 columns C 1 and C 2 . Their product is given by:

$$\begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} 2 & - 2 \\ 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} R_1 \\ R_2 \end{bmatrix} \cdot \begin{bmatrix} C_1 & C_2 \\ \end{bmatrix} = \begin{bmatrix} R_1\cdot C_1 & R_1 \cdot C_2 \\ R_2\cdot C_1 & R_2 \cdot C_2 \end{bmatrix}$$

$$= \begin{bmatrix} \begin{bmatrix} 2 & -1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} & \begin{bmatrix} 2 & -1 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix} \\\\ \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} & \begin{bmatrix} 1 & 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix} \end{bmatrix} = \begin{bmatrix} (2)(2)+(-1)(1)+(0)(-1) & (2)(-2) + (-2)(0) +(0)(2) \\ (1)(2)+(2)(1)+(2)(-1) & (1)(-2) + (2)(0) +(2)(2) \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix}$$

Power of a Matrix

The power of a square matrix A is defined as follows:
$$A^0 = I$$, $$I$$ the
identity matrix
$$A^n = A A .... A$$ (n times) , where n is a positive integer.
If m and n are positive integers, then
$$A^m A^n = A^{m+n}$$
$$(A^m)^n = A^{m n}$$

Properties of Matrix Multiplication

1. The product $$A B$$ of two matrices $$A$$ and $$B$$ is defined if the number of columns of matrix $$A$$ is equal to the number of rows of matrix $$B$$.
2. In general, the product of two matrices is not commutative: $$A B \ne B A$$
3. Matrix multiplication is associative: $$(A B) C = A ( B C)$$ if all the multiplications are defined.
4. Matrix multiplication is distributive: $$A ( B + C ) = A B + A C$$ and $$( A + B ) C = A C + B C$$
5. Multiplication by an identity matrix $$I$$: $$A I = I A = A$$ , this holds for square matrices of dimension n by n.
6. For α and β real: $$\alpha ( A + B ) = \alpha A + \alpha B$$
7. For α and β real: $$\alpha ( \beta A ) = \alpha \beta ( A )$$
8. For α and β real: $$(\alpha + \beta) A = \alpha A + \beta A$$
9. For α real: $$\alpha ( A B ) = (\alpha A) B = A (\alpha B)$$

Questions on Multiplication of Matrices

• Part 1
A, B, C, D and E are matrices with the orders
A: 2 × 3 , B: 3 × 5 , C: 5 × 1 , E: 1 × 5
Which of the following are defined?
1. $$A B$$
2. $$A C$$
3. $$C E$$
4. $$E C$$
5. $$(A B)C$$
• Part 2
A, B, C, D and E are matrices given by: $A = \begin{bmatrix} -1 & 1 & -2 \\ 0 & -2 & 1 \end{bmatrix} ,\quad B = \begin{bmatrix} -1 & 2 & 0 \\ 0 & -3 & 4 \\ -1 & -2 & 3 \end{bmatrix} ,\quad C = \begin{bmatrix} -3 & 2 & 9 & -5 & 7 \end{bmatrix} \\ D = \begin{bmatrix} -2 & 6 \\ -5 & 2 \end{bmatrix} ,\quad E = \begin{bmatrix} 3 \\ 5 \\ -11 \end{bmatrix} ,\quad F = \begin{bmatrix} -1 & 0 & 2 \\ -2 & -3 & 4 \\ 1 & 4 & -3 \end{bmatrix}$ Find if possible:
1. $$A B$$
2. $$B C$$
3. $$A D$$
4. $$E F$$
5. $$F E$$
• Part 3
Find x and y if $\begin{bmatrix} x + y & -2 \\ x - y & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 12 & -8 \\ \end{bmatrix}$
• Part 4
Calculate $\left( \begin{bmatrix} 2&0&0\\ 0&0&2\\ 0&2&0 \end{bmatrix} \right)^{10}$

Solutions to the Above Questions

• Part 1
A, B, C, D and E are matrices with the orders
A: 2 × 3 , B: 3 × 5 , C: 5 × 1 , E: 1 × 5
Which of the following are defined?
1. $$A B$$ : defined because the number of columns of A is equal to the number of rows of B.
2. $$A C$$ : NOT defined, the number of columns of A is NOT equal to the number of rows of C.
3. $$C E$$ : defined because the number of columns of C is equal to the number of rows of E.
4. $$E C$$ : defined because the number of columns of E is equal to the number of rows of C.
5. $$(A B)C$$ : defined because AB is defined (see above) and the results is a matrix of order 2 by 5. The number of columns of AB is equal to 5 which is equal to the number of rows of C.
• Part 2
1. $$A B$$ is defined and is given by
$$A B = \begin{bmatrix} -1 & 1 & -2 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 0 \\ 0 & -3 & 4 \\ -1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 3&-1&-2\\ -1&4&-5 \end{bmatrix}$$
2. $$B C$$ is not defined because the number of columns of B is not equal to the number of rows of C.
3. $$A D$$ is not defined because the number of columns of A is not equal to the number of rows of D.
4. $$E F$$ is not defined because the number of columns of E is not equal to the number of rows of F.
5. $$F E$$ is defined and is given by
$$F E = \begin{bmatrix} -1 & 0 & 2 \\ -2 & -3 & 4 \\ 1 & 4 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \\ -11 \end{bmatrix} = \begin{bmatrix}-25\\ -65\\ 56 \end{bmatrix}$$
• Part 3
Find the product $$\begin{bmatrix} x + y & -2 \\ x - y & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 2x + 2y & -x-y+4 \\ 2x - 2y & -x+y-2 \end{bmatrix}$$
then solve
$$\begin{bmatrix} 2x + 2y & -x-y+4 \\ 2x - 2y & -x+y-2 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 12 & -8 \end{bmatrix}$$

Two matrices are equal if they have the same order and their corresponding entries are equal, hence the system of equations
$$2x + 2y = 8 , -x-y+4 = 0 , 2x - 2y = 12 , -x+y-2 = - 8$$
Solve to obtain x = 5 and y = -1
• Part 4
Calculate Rewrite the matrix as follows: $\begin{bmatrix} 2&0&0\\ 0&0&2\\ 0&2&0 \end{bmatrix} = 2 \begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix}$ Hence $\left( \begin{bmatrix} 2&0&0\\ 0&0&2\\ 0&2&0 \end{bmatrix} \right)^{10} = 2^{10} \left(\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix} \right)^{10}$ We note that the matrix $$\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix}$$ is an Row Operations and elementary matrix corresponding to interchanging rows 2 and 3. So to raise the elementary matrix to the power 10, we start with the elementary and interchange rows 2 and 3 9 times which gives the original matrix $\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$ Hence $\left(\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix} \right)^{10} = 2^{10}\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1024&0&0\\ 0&1024&0\\ 0&0&1024 \end{bmatrix}$