Linear Systems and Matrices PDF

Summary

This document is lecture notes on linear systems and matrices. It covers definitions, operations, and examples of matrices. Basic concepts of matrix algebra are introduced, including addition, subtraction, multiplication, and special types of matrices like null and unit matrices. Examples are provided to illustrate these concepts.

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# **Linear Systems and Matrices** ## **5.1 Definition of a Matrix** - A matrix is a set of *mn* quantities arranged in a rectangular array of *m* rows and *n* columns. - When writing down matrices it is useful to enclose this array by large brackets and denote the matrix by a single letter, say *A...

# **Linear Systems and Matrices** ## **5.1 Definition of a Matrix** - A matrix is a set of *mn* quantities arranged in a rectangular array of *m* rows and *n* columns. - When writing down matrices it is useful to enclose this array by large brackets and denote the matrix by a single letter, say *A*. - This is the general form of a matrix: A = ``` a11 a12 ........ a1n a21 a22 ........ a2n .................... am1 am2 ........ amn ``` - The elements of a matrix, *a<sub>ik</sub>*, are called the elements of the matrix. - A matrix of *m* rows and *n* columns is said to be of order (*m x n*), and to be square if *m* = *n*. ## **5.2 Algebra of Matrices** ### **(a) Addition and Subtraction** - To execute addition and subtraction, matrices must be of the same order. - If *A* and *B* are two matrices of the same order with elements *a<sub>ik</sub>* and *b<sub>ik</sub>*, their sum *A + B* is defined as a matrix *C* whose elements *c<sub>ik</sub> = a<sub>ik</sub> + b<sub>ik</sub>*. - *C* has the same order as *A* and *B*. For example, if: A = ``` (1 2 3) (0 -1 2) ``` and, B = ``` (3 4 5) (1 2 3) ``` Then: A + B = ``` (1+3 2+4 3+5) (0+1 -1+2 2+3) ``` = ``` (4 6 8) (1 1 5) ``` C = A + B = ``` ( 4 6 8) ( 1 1 5) ``` Subtraction of matrices is defined in a similar manner : A - B = ``` (1-3 2-4 3-5) (0-1 -1-2 2-3) ``` = ``` (-2 -2 -2) (-1 -3 -1 ) ``` Also: - A + B = B + A: Associative - (A + B) + C = A + (B + C): Commutative ### **(b) Equality of Matrices** - Two matrices *A* and *B* with elements *a<sub>ik</sub>* and *b<sub>ik</sub>* respectively are equal only if they are of the same order, and if all their corresponding elements are equal (*i.e.* a<sub>ik</sub> = b<sub>ik</sub>). ### **(c) Multiplication by a Number** Example: 3A = ``` (3 * 1 3 * 2 3 * 3) (3 * 2 3 * 3 3 * 4) ``` = ``` ( 3 6 9) ( 6 9 12) ``` ## **5.3 Special Types of Matrices** ### **(a) Null-Matrix** - Any matrix of order (*m x n*) with all its elements equal to zero is called a null matrix of order (*m x n*). N = ``` (0 0) (0 0) ``` ### **(b) Unit matrix** - A square matrix with zero elements everywhere except in the diagonal is a unit matrix. - The general form is: I = ``` (1 0) (0 1) ``` In general: IA = AI = A I = I<sup>2</sup> = I<sup>3</sup> = ...... = I<sup>k</sup> ### **(d) Matrix Multiplication** - The definition of matrix multiplication is such that two matrices *A* and *B* can only be multiplied together to form their product *AB* when the number of columns of *A* is equal to the number of rows of *B*. - Suppose *A* is a matrix of order (*m x p*) with elements *a<sub>ik</sub>*, and *B* is a matrix of order (*p x n*) with elements *b<sub>ik</sub>*. - Then their product *AB* is a matrix *C* of order (*m x n*) with elements *c<sub>ik</sub>* defined by: ``` C<sub>ik</sub> = ∑<sub>s=1</sub><sup>p</sup> a<sub>is</sub> b<sub>sk</sub> ``` For example, if *A* and *B* are (2 x 2) and (2 x 2) matrices respectively given by: A = ``` (a<sub>11</sub> a<sub>12</sub>) (a<sub>21</sub> a<sub>22</sub>) ``` B = ``` (b<sub>11</sub> b<sub>12</sub>) (b<sub>21</sub> b<sub>22</sub>) ``` Then the product C = AB is a (2 x 2) matrix defined as: C = AB = ``` (a<sub>11</sub>b<sub>11</sub> + a<sub>12</sub>b<sub>21</sub> a<sub>11</sub>b<sub>12</sub> + a<sub>12</sub>b<sub>22</sub>) (a<sub>21</sub>b<sub>11</sub> + a<sub>22</sub>b<sub>21</sub> a<sub>21</sub>b<sub>12</sub> + a<sub>22</sub>b<sub>22</sub>) ``` ### **Examples** - A = ``` (1 2) (3 4) ``` B = ``` (2 3) (1 2) ``` Find AB and BA. - AB = ``` (1*2 + 2 * 1 1 * 3 + 2 * 2) (3 * 2 + 4 * 1 3 * 3 + 4 * 2) ``` = ``` (4 7) (10 17) ``` - BA = ``` (2 * 1 + 3 * 3 2 * 2 + 3 * 4) (1 * 1 + 2 * 3 1 * 2 + 2 * 4) ``` = ``` (11 16) (7 10) ``` - A = ``` (2 5) (4 10) ``` B = ``` (3 -1) (1 3) ``` Find AB. - AB = ``` (2 * 3 + 5 * 1 2 * -1 + 5 * 3) (4 * 3 + 10 * 1 4 * -1 + 10 * 3) ``` = ``` (11 13) (22 26) ``` - A = ``` (1 2) (3 4) ``` B = ``` (2 0) (1 1) ``` Find *AB*. - AB = ``` (1*2 + 2 * 1 1 * 0 + 2 * 1) (3 * 2 + 4 * 1 3 * 0 + 4 * 1) ``` = ``` (4 2) (10 4) ``` - A = ``` (-1 2) (3 1) ``` B = ``` (3 -1) (1 -1) ``` Find *AB*. - AB = ``` ( -1 * 3 + 2 * 1 -1 * -1 + 2 * -1) (3 * 3 + 1 * 1 3 * -1 + 1 * -1) ``` = ``` (-1 -1) (10 -4) ``` - Clearly AB ≠ BA even though products are defined. This example shows that matrix multiplication is not commutative. - However, matrix multiplication is commutative in the associative and distributive laws of multiplication: - (AB)C = A(BC): Associative - (A+B)C = AC + BC: Distributive - Provided that the products are defined. - A = ``` (1 2) (0 1) ``` Find A<sup>2</sup>, A<sup>3</sup>, 3A<sup>2</sup> + 3I. - A<sup>2</sup> = A * A = ``` (1 2) * (1 2) (0 1) (0 1) ``` = ``` (1 * 1 + 2 * 0 1 * 2 + 2 * 1) (0 * 1 + 1 * 0 0 * 2 + 1 * 1) ``` = ``` (1 4) (0 1) ``` - A<sup>3</sup> = A * A<sup>2</sup> = ``` (1 2) * (1 4) (0 1) (0 1) ``` = ``` (1 * 1 + 2 * 0 1 * 4 + 2 * 1) (0 * 1 + 1 * 0 0 * 4 + 1 * 1) ``` = ``` (1 6) (0 1) ``` - 3A<sup>2</sup> + 3I = 3 * ``` (1 4) + 3 * ( 0 1) ``` = 3 * ``` (1 4) (0 1) ``` + 3 * ``` (1 0) (0 1) ``` = ``` (3 12) (0 3) ``` + ``` (3 0) (0 3) ``` = ``` (1 - 3+3 0-0+0) (6-12+0 1-3+3) ``` = ``` (1 0) (-6 1) ``` <start_of_image> - A = ``` (-1 2) (2 1) ``` Find (A - 2I)<sup>2</sup>. - A - 2I = ``` (-1 2) - 2 * (2 1) ``` = ``` (-1 2) - (2 0) (0 2) ``` = ``` (-1 2) (-2 1) ``` - (A - 2I)<sup>2</sup> = (A - 2I) * (A - 2I) = ``` ( -1 2) * ( - 1 2) (-2 1) (-2 1) ``` = ``` ( -1 * -1 + 2 * -2 - 1 * 2 + 2 * 1) ( -2 * -1 + 1 * -2 -2 * 2 + 1 * 1) ``` = ``` (-3 0) ( 0 -3) ``` - A = ``` (2 -1) (3 1) ``` B = ``` (1 2) (-1 2) ``` Find A<sup>2</sup>, B<sup>2</sup>, (A+B)<sup>2</sup>, (A-B)<sup>2</sup>, A<sup>2</sup> - B<sup>2</sup>, A<sup>2</sup> + B<sup>2</sup>, A<sup>3</sup> - 2A<sup>2</sup> + 3I, B<sup>4</sup> + 3B<sup>2</sup> - 5I. - A<sup>2</sup> = A * A = ``` (2 -1) * (2 -1) (3 1) (3 1) ``` = ``` (2 * 2 + -1 * 3 2 * -1 + -1 * 1) ( 3 * 2 + 1 * 3 3 * -1 + 1 * 1) ``` = ``` (1 -3) (9 -2) ``` - B<sup>2</sup> = B * B = ``` (1 2) * (1 2) (-1 2) (-1 2) ``` = ``` (1*1 + 2 * -1 1 * 2 + 2 * 2) (-1 * 1 + 2 * -1 -1 * 2 + 2 * 2) ``` = ``` (-1 6) (-3 2) ``` - A + B = ``` (2 -1) + (1 2) (3 1) (-1 2) ``` = ``` (2+1 -1+2) (3-1 1+2) ``` = ``` (3 1) (2 3) ``` - (A + B)<sup>2</sup> = (A + B) * (A + B) = ``` (3 1) * (3 1) (2 3) (2 3) ``` = ``` (3 * 3 + 1 * 2 3 * 1 + 1 * 3) (2 * 3 + 3 * 2 2 * 1 + 3 * 3) ``` = ``` (11 6) (12 11) ``` - A<sup>3</sup> = A. A<sup>2</sup> = ``` (2 -1) * (1 -3) (3 1) (9 - 2) ``` = ``` (2*1 + -1 *9 2*-3 + -1 * -2 ) (3*1 +1 * 9 3 * -3 + 1 * -2) ``` = ``` (-7 4) (12 -11) ``` - 2A<sup>2</sup> = 2 * ``` (1 -3) (9 -2) ``` = ``` (2 -6) (18 -4) ``` - 3I = 3 * ``` (1 0) (0 1) ``` = ``` (3 0) (0 3) ``` - A<sup>3</sup> - 2A<sup>2</sup> + 3I = ``` (-7 4) ( 12 -11) ``` - ``` ( 2 -6) (18 -4) ``` + ``` (3 0) (0 3) ``` = ``` (-7 - 2 + 3 4 - 6 + 0) (12 - 18 + 0 -11 + 4 + 3) ``` = ``` (-6 2) (-6 -4) ``` - A = ``` (2 -1) (3 1) ``` - A<sup>2</sup> = ``` (1 -3) (9 -2) ``` - A<sup>3</sup> = A * A<sup>2</sup> = ``` ( 2 -1) * (1 -3) ( 3 1) (9 -2) ``` = ``` (2 * 1 + -1 * 9 2 * -3 + -1 * -2) (3 * 1 + 1 * 9 3 * -3 + 1 * -2) ``` = ``` (-7 4) (12 -11) ```

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