Chapter 1 Linear Equations and Matrices

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1 Chapter : Linear Equations and Matrices Chapter Linear Equations and Matrices SECION A Systems of Linear Equations By the end of this section you will be able to understand what is meant by a linear system solve a linear system consisting of two equations with two unknowns solve a linear system consisting of three equations with three unknowns plot linear graphs and determine the type of solution Many problems in engineering and science are solved by using systems of simultaneous linear equations. For example, if you apply Hooke s law to a spring then this leads to a system of linear equations. If you apply Kirchhoff s law to a circuit which only contains resistors then the result is a set of linear equations. Linear equations arise in many different fields such as engineering, physics, computer science, technology, business, economics etc. Generally the complex scientific or engineering problem can be reduced to a linear system. In this section you will need to know how to solve simultaneous equations and plot straight line graphs. It is not a difficult section but make sure you are familiar with these topics. If you are not comfortable in solving simultaneous equations and plotting graphs then revise these topics before you embark on your journey to understand this section. A Introduction to Linear Equations What does the term linear equation mean? An equation is where two mathematical statements are equal. A linear equation is an equation where all the variables such as x, y, z etc have index (power) of or 0 only, for example x+ y+ z =5 is a linear equation. he following are linear equations: ) x + y+ z = ) x+ y = 5 ) x+ y+ z+ w= 8 he following are not linear equations: 4) x = 0 5) x y = 8 4 6) x+ y + z = 9 7) sin ( x) y+ z = Why not? In equation 4) the index (power) of the variable x is and this is actually a quadratic equation. In equation 5) the index of x is and this is again a quadratic equation. In equation 6) the index of y is 4 and z is /. emember z = z. In equation 7) the variable x is an argument of the trigonometric function sine. Notice that linear equations do not contain squares, cubes, square roots, cube roots etc of the unknown variables. Also if the unknown variable such as x is an argument of trigonometric, exponential, logarithmic or hyperbolic functions in an equation then the equation is not linear.

2 Chapter : Linear Equations and Matrices We can plot linear equations on a graph. he following is an example of three linear equations on a graph. y x =0 y y x = x - y x +=0 - Fig Here is another example of linear equations. 0 y - y +5x = 5 y x = x -5-0 Fig What do you notice about the graphs of linear equations? hey are all straight lines and this is why they are called linear equations and the study of such equations is called linear algebra. A System of Linear Equations What does the term system of linear equations mean? wo linear simultaneous equations are an example of a system of linear equations. For example the two equations in Fig, y + 5x = y x = is an example of a system of linear equations. Generally a finite number of linear equations with a finite number of unknowns x, y, z, w, L is called a system of linear equations or just a linear system. For example the following is a linear system of three simultaneous equations with three unknowns x, y and z: x + y z = x y z = x + y + z = 5

3 Chapter : Linear Equations and Matrices In general a linear system of m equations in n unknowns x, x, x, L and xn is given by ax + ax + L + a nxn = b ax + ax + L + anxn = b ax + ax + L + anxn = b ( ) M M M M am x + amx + L + amnxn = bm where the coefficients a ij 's and b j 's are real numbers. he unknowns x, x, L and xn are place-holders for real numbers. Linear algebra involves finding solutions of linear systems such as ( ). Example Solve the above equations of graphs in Fig by algebraic means: y + 5x = (*) y x = (**) Solution How do we solve these linear simultaneous equations, (*) and (**)? You must have covered simultaneous equations at GCSE or A level but can you remember the procedure used to solve these? By elimination. We eliminate one unknown, x or y, and then solve one equation with the other unknown. How? Subtract the equations (*) and (**): y + 5x = (*) ( y x = ) (**) 0 + 7x = 4 Note that the unknown y is eliminated in the last line and from this we have 7 x = 4 which gives x =. What else do we need to find? he other unknown y. How? By substituting x = into (*). y + ( 5 ) = y + 0 = gives y = Hence x = and y = is the solution to the given simultaneous equations (*) and (**). his is also the point of intersection, (, ), of the graphs in Fig. he procedure outlined in Example is called the method of elimination. he values x = and y = is the solution of (*) and (**). In general numbers such as s, s, s, L, sn which satisfy the above linear system ( ) is called the solution or the solution set of the linear system. What does this mean? Means that if we substitute x = s, x = s, x = s, L and xn = sn then every equation in ( ) is satisfied. Here is another example. Example Solve 9x + y = 6 ( ) x 7y = 9 ( )

4 Chapter : Linear Equations and Matrices 4 Solution How do we solve these linear simultaneous equations, ( ) and ( )? o eliminate one of these unknowns, you need to make the coefficients of x (or y) equal. Which one? It doesn t really make much difference apart from easier arithmetic. If we multiply ( ) by and ( ) by 9 then in both cases the x coefficient is 8. Carrying out this operation we have 8x + 6y = Multiplying ( ) by 8x 6y = 8 Multiplying ( ) by 9 How do we eliminate x from these equations? o eliminate the unknown x we subtract these equations: 8x + 6y = (8x 6y = 8) ( ) y [ ] = 8 Subtracting 69y = 69 which gives y = We have y =. What else do we need to find? he value of the place-holder x. How? By substituting y = into the given equation 9x + y = 6 : ( ) 9x + = 6 9x = 6 9x = 9 which gives x = Hence our solution to the linear system of ( ) and ( ) is x = and y = We can check that this is the solution to the given system, ( ) and ( ), by substituting these values, x = and y =, into the equations ( ) and ( ): () ( ) () ( ) ( ) ( ) ( ) 9+ = 9 = 6 9x + y = 6 7 = + 7= 9 x 7y = 9 Note that we can carry out the following operations on a linear system of equations:. Interchange any pair of equations.. Multiply an equation by a nonzero constant.. Add or subtract one equation from another. here is nothing new here, you have carried out these three steps when you solved simultaneous equations at GCSE or A level. By carrying out these steps, and we end up with another linear system but with the same solution set as the given linear system. We can also use this method of elimination to solve three simultaneous linear equations with three unknowns, such as the one in the next example. Example Solve the linear system x + y + 4z = 7 (*) x + 7y + z = (**) x + y + z = (***) Solution What are we trying to find? he values of x, y and z that satisfy all three equations (*), (**) and (***).

5 Chapter : Linear Equations and Matrices 5 How do we find the values of x, y and z? By elimination. o eliminate one of these unknowns, you need to make the coefficients of x (or y or z) equal. Which one? here are many choices but we select so that our arithmetic is easier and in this case it is x. Multiply the above given equation (*) by and then subtract the bottom equation (***): x + 4y + 8z = 4 Multiplying (*) by [ ] ( x 0 + y y + z 5z = ) (***) [ Subtracting] + + = Note that we have eliminated x and have the equation y+ 5z =. How can we determine the values of y and z from this equation? We need another equation with only y and z unknowns. How can we get another equation with the place-holders y and z only? Multiply the given equation (*) by and then subtract the second equation (**): x + 6y + z = Multiplying (*) by [ ] ( x 0 7y y + z 0z = ) (**) [ Subtracting] + + = Again there is no x and we have the equation y+ 0z =. How can we find the values of the unknowns y and z? We now solve the two simultaneous equations that we have obtained: y + 5z = ( ) y + 0z = ( ) How do we solve these? We add these, ( ) and ( ), equations. Why? Because y+ y = and so adding eliminates y. ( ) z = 45 [Adding ( ) and ( )] z = 45 5 = Hence z = but how do we find the other two unknowns x and y? We first determine y by substituting z = into the above equation y + 5z = ( ): y ( ) + 5 = y + 5 = which gives y = We have y = and z =. What else do we need to find? he value of x because we have three unknowns x, y and z and we have already found y = and z =. How do we find the value of x? By substituting the values we have already found, y = and z =, into the given equation x + y + 4z = 7 * : ( ) ( ) ( ) x = 7 x 4 + = 7 x = = Hence the solution of the given three linear equations is x =, y = and z =. You may like to check that this indeed is the solution of the given system. We will examine in detail m equations with n unknowns in the next few sections and develop

6 Chapter : Linear Equations and Matrices 6 a more efficient way of solving these. A ypes of Solutions We now go back to two linear simultaneous equations and discuss the case where we have no or an infinite number of solutions. Example 4 Solve the linear system x + y = 6 ( ) 4x + 6y = 9 ( ) Solution How do we solve these equations? Multiply ( ) by and then subtract ( ): 4x + 6y = Multiplying ( ) by 4x + 6y = 9 [ ] ( ) ( ) = But how can we have 0 =? his is impossible. What mistake have we made? here is no mistake but if we plot the graphs of the given equations we have: y hese equations have no solutions. 4x +6y =9 x +y =6-4 x - Fig Can you now see why there is no solution to these equations? he solution is the intersection of the lines of the given equations shown in Fig but these lines are parallel so they do not intersect therefore there is no solution. By examining the given equations x + y = 6 ( ) 4x + 6y = 9 ( ) can you see why there is no solution? If you multiply the first equation ( ) by we have 4x + 6y = his is a contradiction. Why? Because we have 4x + 6y = 4x + 6y = 9 ( )

7 Chapter : Linear Equations and Matrices 7 that is 4x + 6y equals both 9 and. his is clearly impossible. Hence the given linear system has no solution. A system that has no solution is called inconsistent. If the linear system has at least one solution then we say the system is consistent. Can we have more than one solution? Consider the following example. Example 4 Graph the equations: x + y = 6 ( ) 4x + 6y = ( ) Determine the solution of this system. Solution Graphing the given equations we have: 4 y x +y =6 4x +6y = x - Fig 4 What do you notice? Both the given equations produce the same line that is they coincide. How many solutions do these equations have? An infinite number of solutions as you can see on the graph of Fig 4. Any point on the line is a solution and since there are an infinite number of points on the line therefore we have an infinite number of solutions. But how can we show this by algebraic means? Let x = a where a is any real number, be a solution. What is y equal to? Substituting x a yields Hence if x x = a and = into the given equation ( ) = a then ( ) a + y = 6 x + y = 6 y = 6 a 6 a 6 y = = a = a y = a. he solution of the given linear system, ( ) and ( ), is y = a where a is any real number. You can check this by substituting various a = then values of a. For example let 4 x=, y = () =

8 Chapter : Linear Equations and Matrices 8 We can check that this answer is correct by substituting these values, the given equations ( ) and ( ): 4 () + = + 4= () + 6 = 4+ 8= Hence our solution works. his solution x 4 x= and y =, into ( ) x + y = 6 4x + 6y = ( ) = a and equations for any real value of a. ry other values of a such as and check that these values, x and ( ). = a and y = a will satisfy the given 0,, π,, e, 00,, L y = a, satisfy the given linear equations, ( ) here are three types of solutions to linear systems. a) No solution. b) One solution. c) Infinite number of solutions. hese correspond to the following graphs of two linear equations with two unknowns: y y x -4-4 x Solution (a) No Solution (b) One Solution y 0 5 wo equations on the same line x -5 Fig 5-0 (c) Infinite number of Solutions

9 Chapter : Linear Equations and Matrices 9 A4 Equivalent Systems Consider the general linear system: ax + ax + L + a nxn = b ax + ax + L + anxn = b ( ) M M M M am x + amx + L + amnxn = bm Earlier we stated that the following operations can be executed on a linear system:. Interchange any pair of equations.. Multiply an equation by a nonzero constant.. Add or subtract one equation from another. We also claimed that as a result of these operations we end up with the same solution set. How do we know this? Let x = s, x = s, L, xn = sn be the solution of this given linear system ( ).. If we carry out the first operation of changing a pair of equations then we result in the same list therefore we have the same solutions.. Multiplying an equation, say the rth equation, ars + ars + L + arnsn = bn (*) by a nonzero constant k gives kars + kars + L + karnsn = kbn (**) Hence x = s, x = s, L, xn = sn is also the solution of k times the original equation. his means that x = s, x = s, L, xn = sn is a solution of (*) if and only if it is a solution of (**).. Suppose we add rth and pth equations: ars + ars + L + arnsn = br + a s + a s + L + a s = b p p pn n a s + a s + a s + a s + L + a s + a s = b + b r p r p rn n pn n r Factorising gives ( ) ( ) L ( ) ar + ap s + ar + ap s + + arn + apn sn = bn + bp his means that x = s, x = s, L, xn = sn is also a solution of adding two equations. We can say that if x = s, x = s, L, xn = sn is the solution of the above system ( ) then it is also the solution of any version of the above system which has been changed by the above operations, and. SUMMAY A linear equation is an equation where the unknowns (place-holders) have an index of or 0. A linear set of m equations with n unknowns is called a linear system. Linear Algebra is the study of such linear systems. he procedure outlined in this section to solve such systems is the method of elimination. A solution to a system is a sequence of real numbers s, s, s, L, sn which satisfy the given linear system. here are three types of solutions to a linear system: a) No solution. b) One solution. c) Infinite number of solutions. p p

10 Chapter : Linear Equations and Matrices 0 Exercise (a). Which of the following equations are linear equations in x, y and z? If they are not linear explain why not. (a) x y z = (b) x y z 6 cos x + sin y = (d) + + = (c) ( ) ( ) x y z e + + = (e) x y+ 5z = (f) x = y b± b 4ac (g) x = (h) π x+ y+ ez = 5 (i) x+ y+ z = 0 a π (j) sinh ( x) = ln x+ x + (k) sin( ) x y + z π = (l) x + y + z = 0 (m) y ( x) + sin ( x) cos + x z = 9. Solve the following linear system by the elimination process discussed in this section. (a) x x + y = y = 0 (b) x y = x y = 5 (c) x y = 5 x y = 5x 7y = π x 5y = (d) (e) 9x y = 6 π x y = [In part (f) e is the irrational number e = L ] (f) ex ex ey = + ey = 0. Solve the following linear system by the elimination process discussed in this section. x + y + z = x + y z = 6 (a) x y z = (b) x y + z = 0 x + y + 5z = 8 x y + z = 6 (c) x + y z = 4 5x y + 0z = 7x + 4y + 6z = (d) 6x y + z = 5x + y + z = 8x + 5y + z = 6 4. Plot the graphs of the following linear equations and decide on the number of solutions of each linear system. If there are any solutions find them. x + y = x + y = (a) (b) (c) x + y = x y = 7 8x + 4y = x + y = 5 (d) x y = x y = 6 (e) x y = x y 5 = 0 0 (f) 5 x y 5 = x y = 0 0

11 Chapter : Linear Equations and Matrices Brief Solutions to Exercise a. (a) Linear (b) Not Linear (c) Not linear (d) Not Linear (e) Linear (f) Linear (g) Linear (h) Linear (i) Linear (j) Not Linear (k) Linear (l) Linear (m) Linear (n) Not Linear. (a) x =, y = (b) x =, y = (c) x = 9, y = (d) x=, y = (e) x =, y = (f) x, y 4 4 4π 4 e e. (a) x = y = z = (b) x =, y = and z = (c) x=, y = 4 and z = (d) x =, y = and z = 4. (a) One solution (b) Infinite number of solutions 0 y 6 y x = a, y = a 5 x + y = 4 8x +4y = x x + y = -5 x y = x -0 Point of Intersection (Solution) ( 0/, -/ ) - (c) No solution 6 y (d) No solution 6 y 4 x + y =5 4 x y = x + y = x y =6-4 x x - - (e) No solution y 6 (f) One solution y 6 4 x y =0 4 5x y 5=0 x y 5=0 x y =0-4 5 x - x - - Point of Intersection (, 0 )

12 Chapter : Linear Equations and Matrices SECION B Arithmetic of Matrices By the end of this section you will be able to understand what is meant by a matrix add and subtract matrices evaluate scalar multiplication multiply matrices What does the term matrix mean? A matrix (plural matrices) is an array of numbers enclosed in a bracket Matrices are used in various fields of engineering, science and economics to solve real life problems such as those found in control theory, electrical principles, structural analysis, string theory, quantitative finance, economic models and many others. Problems in these areas can be written in terms of linear simultaneous equations and in most cases it is easier to solve these by converting the equations to matrix form and then use matrix theory rather than employing elimination as was used in the last section. Another application of matrices is in image processing. Many displays form images by lighting up tiny dots, called pixels, on their screens. Figure 6 shows a symbol represented by lighting up appropriate pixels. Fig 6 If we let = pixel on and 0= pixel off then we can represent this in matrix form as his is an example of a 4 5 (4 rows by 5 columns) matrix. he numbers in the array are called the entries or the elements of the matrix. B Introduction to Matrices Fig 7 Arthur Cayley (8 to 895) was an English lawyer who become a mathematician and was the first person to develop matrices as we know them today. He graduated from rinity College, Cambridge in 84 and taught there for four years. In 849 he took up the profession of lawyer and worked at the courts of Lincoln s Inn in London for 4 years. He returned to mathematics in 86 and was appointed Professor of Pure Mathematics at Cambridge with a substantial decrease in salary but was very happy to pursue his interest in mathematics. In 858 he published Memoir on the heory Of Matrices which contained his definition of a matrix.

13 Chapter : Linear Equations and Matrices Cayley thought that matrices were of no practical use, just a convenient means of notation. He could not have been more wrong. Matrix theory or linear algebra is now fundamental to engineering, sciences, medicine, statistics, economics etc. he size of the matrix is determined by the number of rows times the number of columns. 6 For example is a matrix ( rows and columns) and is called a square 7 matrix. his is verbally stated as by matrix. We call this (or by ) the size of the matrix. 7 6 is a 4 matrix (4 rows and columns) and is not a square matrix. An example of a 4 matrix is. he size of this matrix is verbally stated as by 4 matrix Note that when we say 4 matrix we state the number of rows first and then the number of columns. Hence 4 and 4 are different size matrices. A common notation for matrices is: Column Column where element Element a a a ow a a ow a a is the entry in the first row and second column. What row and column is the in? is in the second row and first column. Note that the subscript of each element states row number first and then the column number. he position of different from. What is the position of an element a in a 4 matrix? Element a a is in the second row and third column of a by 4 matrix: Col. a in a matrix is a a a a4 ow a a a a4 = A a a a a4 Generally throughout this chapter bold capital letters will represent matrices and lower case letters the elements (or entries) of the matrices. B Addition and Subtraction of Matrices o add or subtract matrices, we add or subtract the corresponding locations in each matrix respectively. For example = = Also = =

14 Chapter : Linear Equations and Matrices 4 Note that matrices must be of the same size in order to add or subtract. In general let A and B be matrices of size m n ( m rows by n columns) given by a L a n b L b n A = M M M and B = M M M am a L m n bm b L m n he lower case letters are the elements (or the entries) of the matrix and their position in the matrix is given by their subscript. he sum of these matrices is defined by a L a n b L b n a + b L a n + b n (.) A+ B= M M M + M M M = M M M a m a mn bm b L L mn am+ bm L amn + bmn Similarly the subtraction of these matrices is defined by a L a n b L b n a b L a n b n (.) A B= M M M M M M = M M M a m a mn bm b L L mn am bm L amn bmn Note that for addition (or subtraction) we add (or subtract) the elements in the corresponding locations and result of this is another matrix of the same size as the initial matrices. In the matrix A+ B what element is placed in the second row and third column? a + b In the matrix A B what element is placed in the third row and first column? a b Example 5 Determine the following: (a) + (b) (c) (d) (e) Solution (a) Adding the corresponding elements of each matrix gives = = (b) Again adding the corresponding entries of the given matrices we have = = (c) Subtracting the corresponding elements of the matrices gives = = (d) Similarly we have

15 Chapter : Linear Equations and Matrices = = (e) Cannot evaluate the given matrices. Why not? Because they are of different size. he first matrix,, is a 5 4 matrix and the second 4 matrix,, is a matrix therefore we cannot subtract (or add) these matrices emember when we add or subtract matrices the resulting matrix is always the same size as the given matrices. B Scalar Multiplication of Matrices What does the term scalar mean? A scalar is a number which affects the magnitude of the matrix. What does scalar multiplication mean? Let A be a matrix and k be a scalar then the scalar multiplication ka is the matrix constructed by multiplying each entry of A by k. a L a n For example let A = M M M then am a L m n ka L kan (.) ka = M M M kam ka L mn Note that scalar multiplication of a matrix gives another matrix of the same size. Scalars are real numbers but may also be complex numbers. Example 6 Determine the following: 5 7 (a) (b) (c) Solution (a) Multiplying each entry of the matrix,,, and 4, by 0 gives = = (b) Multiplying each element in the second matrix by ½ (or dividing by ) gives

16 Chapter : Linear Equations and Matrices = = (c) Multiplying each element of the matrix by 7 gives = = We can also go the other way, that is factorize out a common factor. If we have the matrix ka, we can take out the common factor k from every entry of the matrix. For example = = Hence we have factored out 5 because each entry of the matrix is a multiple of 5. B4 Multiplication of Matrices he following is not how you multiply one matrix by another: [ Not Equal] X Matrix multiplication does not follow the same rules as matrix addition and subtraction. Matrix multiplication is carried out by row times column: Col. Col. ow 4 9 ow Col. ow Col. = ow ow Col. ow Col. Matrix multiplication is row column and the formal rule is: a b e f ae + bg af + bh = c d g h ce + dg cf + dh Applying this to the above matrices gives 4 9 ( 4) + ( 5) ( 9) + ( 8) = ( 4) + ( 7 5) ( 9) + ( 7 8) 6 = 47 8 We can only multiply matrices if the number of columns of the first (Left Hand) matrix equals the number of rows of the second (ight Hand) matrix. In general if A is a m r (m rows by r columns) matrix and B is a r n (r rows by n columns) matrix then the multiplication AB is a m n matrix. he following formula for matrix multiplication might seem to be confusing but there is no

17 Chapter : Linear Equations and Matrices 7 easy way of stating this. Let Col. Col. Col. r Col. Col. Col. n a a L a r b b L b n ow A = M M M M and B = M M M M M a m a mr br b L L rn ow r then matrix multiplication AB is a matrix of size m n and defined by (.4) ( a b ) + L+ ( a r br ) L L ( a b n) + L+ ( a r brn) AB = M M M M ( am b) + L+ ( amr br) L ( am bn) + + ( amr brn) L L Clearly this is an horrendous formula to remember and it is easier to recall that matrix multiplication is row times column. Let s do some examples. Example 7 Compute the following, if possible: (a) (b) ( 4) (c) ( 0) (d) (e) ( (f) ) Solution (a) Using the concept of row times column we have ( ) + ( ) 7 = = 5 ( ) ( 5 + ) (b) Similarly we have ( 4) = (( ) + ( 4 ( ) ) ( ) + ( 4 5) ) 5 = 9 = ( + ) = ( ) = 4 ( ) (c) ( 0) ( ) ( 0 4). Note that we can write a matrix such as () without brackets as. (d) We have 7 6 ( ) + ( 4) + ( 6 ) ( 7) + ( ) + ( 6 ) = ( ) ( 5 4) ( 7 ) ( 7) ( 5 ) ( 7 ) = 0 8 (e) ( 4 ) 5 = (( 4) + ( 5) + ( 6) ) = ( ) =. As in case (c) we can write the 6 matrix () without brackets as.

18 Chapter : Linear Equations and Matrices (f). Since the number of columns in the Left Hand matrix is and the 5 74 number of rows in the ight Hand matrix is we cannot multiply these matrices. emember these must be equal. hat is the number of columns in the Left Hand matrix must equal the number of rows in the ight Hand matrix. A matrix with just one row such as ( ) is called a row matrix. In general a matrix of size n is a row matrix. Are there any other row matrices in Example 7? Yes, and 0. What do you think the term column matrix means? ( 4) ( ) A matrix with just one column such as. A matrix of size n is a column matrix. Are there any other column matrices in Example 7? 4 Yes, and 5. Is a matrix a column or row matrix? 4 6 It is both a row and column matrix. We can write the computation of matrices in compact form. For example let A and B be matrices then A+ B means add the matrices A and B. he computation A means multiply every entry of the matrix A by. A software package such as MALAB can be used for computing matrices. In fact MALAB is short for Matrix Laboratory. It is a very useful tool which can be used to eliminate the drudgery from lengthy calculations. here are many other mathematical software packages available but MALAB is particularly useful for linear algebra. MALAB commands and other details are given on the website. In the present release of MALAB the matrices are entered with square brackets rather than round ones. For example, using MALAB to write the matrix A = we enter A=[ ; 4] after 4 the command prompt >>. he semicolon indicates the end of the row. he output shows: A = 4 Example 8 Let A=, B= 7 and = C Compute the following: (a) B+ C (b) B+C (c) A+B (d) B 5C (e) AB (f) BA What do you notice about your results to parts (e) and (f)? (a) Adding the corresponding entries gives B+ C= = = o use MALAB enter the matrices B and C after the command prompt >>. Separate each matrix by a comma. hen enter B+ C and the output should give the same result as above.

19 Chapter : Linear Equations and Matrices 9 (b) We have B+ C= Multiplying each = entry of matrix by + B and matrix C by Simplifying each = entry Adding the = corresponding = entries In MALAB, multiplication is carried out by using the command *. Once the matrices B and C have been entered you do not need to enter them again. Enter * B+ * C for checking the above evaluation. (c) A+B is impossible because matrices are of different size. What size is matrix A and matrix B? A = is a matrix and B = 7 is a matrix therefore we cannot 9 6 add (or subtract) these matrices. If we try this in MALAB we receive a message saying Matrix dimensions must agree. (d) B 5C is evaluated by B 5C= Multiplying each = entry of matrix C by = = =

20 Chapter : Linear Equations and Matrices 0 (e) We have AB = ( ) + ( ) + ( 5 9) ( 5) + ( 7) + ( 5 6) ( 6) + ( ) + ( 5 ) ( 4 ) + ( 6 ) + ( 9 9) ( 4 5) + ( 6 7) + ( 9 6) ( 4 6) + ( 6 ) + ( 9 ) = = [ Simplifying] = (f) BA is impossible because the number of columns in the Left Hand matrix, B, is not equal to the number of rows in the ight Hand matrix, A. How many columns does the matrix B have? 5 6 B = 7 has columns. How many rows does the matrix A have? Matrix A = has rows. Since the number of columns of matrix B,, does not match the number of rows of matrix A,, therefore we cannot evaluate BA. If we try this in MALAB we receive an error message saying Inner matrix dimensions must agree. A fundamental difference between ordinary and matrix algebra is that the matrix multiplication AB does not equal the matrix multiplication BA. In fact we have established in this case that BA cannot be evaluated whilst AB is computed in part (e) above. It is important to note that matrix multiplication is not the same as multiplying two real numbers. In matrix multiplication the order of the multiplication does matter. We will discuss important arithmetic properties of matrices in the next section. SUMMAY A matrix is an array of numbers. We can add or subtract matrices of the same size. Scalar multiplication affects the magnitude of the matrix and is defined by a L a n ka L ka n (.) k M M M = M M M am a L mn kam L ka mn Matrix multiplication is row times column. Matrix multiplication AB is only valid if the number of columns in the Left Hand matrix, A, matches the number of rows in the ight Hand matrix, B.

21 Chapter : Linear Equations and Matrices Exercise (b) You may like to check your numerical solutions by using MALAB. 6. For =, = and = 5, evaluate, where possible, the following: (a) A+B (b) B+A (c) B+B+B (d) B (e) A+ B (f) A+C (g) B+ C (h) AC (i) BC (j) 5A 7BC (k) AC BC What do you notice about your answers to parts (c) and (d)?. Consider the following matrices: A=, B= 8 7, 4 and C= D= Compute the following where possible: (a) A A (b) A A (c) BC (d) CB (e) B+ D (f) D+ B (g) A C (h) C (i) BD (j) DB (k) BD DB (l) CD. Evaluate the following: (a) (b) (c) 90 0 where a, b, c and d are any real numbers (d) What do you notice about your results? a c b 0 d0 4. Determine the following: (a) 5 (b) (c) Let A = and find A = A A. [ A in MALAB is evaluated by the command A ^]. Comment upon your result. Can you get the same result when multiplying two non-zero real numbers? 5 6. Evaluate 0 4. Comment upon your result. Can you get the same result when multiplying two non-zero real numbers?

22 Chapter : Linear Equations and Matrices 7. Complete the subscript numbers in the following matrix: a a a a a a a a 8. ake out a common factor from the following matrices: (a) (b) 4 (c) (d) Let A = 0. Determine 0 A = A A and A = A A A Let A =. Find 0 4 A = A A, A = A A A and A = A A A A. Brief Solutions to Exercise (b) (a) (b) (c) (d) (e) (h) (i) (k) Parts (f), (g) and (j) cannot be evaluated. Parts (c) and (d) are the same because B+ B+ B=B (a) (b) = A (c) 7 (d) Impossible (e) 9 5 (f) 9 5 (g) Impossible (h) (i) 7 7 (j) 4 0 (k) (l) Impossible a b. (a) (b) (c) (d) c d he result is always the first (or Left Hand) matrix (a) (b) (c) 0 0 0

23 Chapter : Linear Equations and Matrices Multiplying two non-zero matrices gives a zero matrix. his cannot occur 0 0 when multiplying two real numbers Multiplying two non-zero matrices gives a zero matrix. his cannot occur when multiplying two real numbers. a a a a 4 7. a a a a (a) 6 (b) (c) (d) A =, A = A =, =, = 0 A 0 A 0

24 Chapter : Linear Equations and Matrices 4 SECION C Properties of Matrix Algebra By the end of this section you will be able to simplify matrices by using algebra prove and use results of matrix addition, scalar multiplication and matrix multiplication compare and contrast between matrix and real number algebra his section is more difficult then previous sections because it requires you to prove results and understand various properties of matrix algebra. First we discuss the zero matrix. What do you think the term zero matrix means? A zero matrix is a matrix which has all zero entries and it is denoted by O. he, 4 and zero matrices are , and respectively Sometimes the zero matrix is denoted by O meaning it is a zero matrix of size m n (m mn rows by n columns). he above matrices are denoted O, O4 and O respectively. C Properties of Matrix Addition In Exercise b question parts (a) and (b) we showed for particular matrices A and B that A+ B = B+ A his is true for all matrices that can be added. In Exercise b question parts (c) and (d) we also showed for particular matrices B+ B+ B=B. In general B4 + B B+ L+ B=nB n copies Most of the laws of algebra of real numbers can also be extended to algebra of matrices. hese laws and the above properties are summarised in the next theorem. heorem (.5). Properties of Matrix Addition. Let A, B and C be matrices of the same size m n (m rows by n columns). hen (a) A+ B = B+ A (b) ( A+ B) + C= A+ ( B+ C) (Associative law for addition) (c) A+ A+ A+ L+ A =ka k copies (d) here is a m nmatrix called the zero matrix denoted O which has the property A+ O= A (e) here is a matrix denoted A such that A+ A = A A= O ( ) his matrix A is called the additive inverse of A. Proof. We will prove parts (a), (c) and (d) only. he rest are left as questions in Exercise (c). a L a n b L b n Let A = M M M and B = M M M. am a L m n bm b L m n

25 Chapter : Linear Equations and Matrices 5 (a) emember the order of adding real numbers does not matter, that is for real numbers a and b we have a+ b = b+ a. Adding the matrices A and B given above yields a L a n b L b n a + b L a n + b n Adding the A+ B= corresponding M M M + M M M = M M M am a mn bm b mn am bm amn b L L + L + mn entries b + a L b n + a n Changing the = M M M Order of bm + am bmn + a L mn Addition b L b n a L a n = M M M + M M M bm b mn am a L L mn = B+ A Hence we have shown our required result, A+ B = B+ A. (c) Need to prove A+ A+ A+ L+ A =ka. We have k copies = B = A a L a n a L a n a L an A+ A+ A+ L+ A= M M M + M M M + L+ M M M k copies a L a a L a a L a We have proven A+ A+ A+ L+ A =ka. k copies m mn m mn m mn a + a + L+ a L a n + a n + L+ an k copies k copies = M M M am + am + L+ am L amn + amn + L+ a mn k copies k copies ka L ka n a L a n = M M M = k{ M M M = ka Factorizing kam ka mn Out k am a L L mn 4444 = A (d) equired to prove A+ O= A where O is the zero matrix. Writing the matrices A and O: a L a n 0 L 0 A+ O= M M M + M M M am a mn 0 0 L L a + 0 L an + 0 a L a n = M M M = M M M = A am + 0 amn + 0 L am L a mn Hence A+ O= A.

26 Chapter : Linear Equations and Matrices 6 he symbol signifies the end of the proof. We will use this symbol throughout the book. We can use these results to simplify matrices as the next example shows. Example 9 Let A =, = and =. Determine B 4 5 C 9 (a) A+ B (b) A+ B+ C (e) A+ O (f) Solution (a) We have A+ B= B A (c) ( A B) C (d) ( ) A A A A A (g) C+ ( C) ( ) = = (b) Clearly by the above heorem (.5) (a) we have B+ A= A+ B = [By part (a)] 0 (c) We have already evaluated A+ B in part (a). herefore we have ( A+ B) + C= (d) (e) (f) (g) By Part (a) = = 6 4 By heorem (.5) (b) we have A+ ( B+ C) = ( A+ B) + C = { By Part (c) 6 4 By heorem (.5) (d) we have 7 A+ O= A = By heorem (.5) (c) we have 7 A A+ A+ A+ A= 5A= 5 = 5A Multiplying each = entry by = [ Simplifying] By heorem (.5) (e) we have C+ ( C) = C C= = O

27 Chapter : Linear Equations and Matrices 7 C Properties of Scalar Multiplication heorem (.6). Let A and B both be m n(m rows by n columns) matrices and c, k be scalars. hen ck A= c ka (a) ( ) ( ) (b) k( ) k (c) ( ) A+ B = A+ kb c+ k A= ca+ ka (d) ka= Ak a L a n Proof. We only prove part (c). Let A = M M M then am a L m n a L a n ( c+ k) A = ( c+ k) M M M am a L mn ( + ) L ( + ) c k a c k a n Multiplying each entry = M M M by ( c k) ( c k) am ( c k) a + + L + mn ca + ka L ca n + ka n Opening up the = M M M Brackets cam + kam camn + ka L mn ca L ca n ka L ka n Separating Out = M M M + M M M erms into c's and k's cam ca mn kam ka L L mn a L an a L a n = c{ M M M + k{ M M M = ca+ ka Factorizing Factorizing Out c am a mn Out k am a L L mn = A = A Example 0 Let A =, =, c= and 5. Determine the following: 9 6 B k = k A+ B (b) k + k c+ k B (d) cb+ k B (a) ( ) A B (c) ( ) (e) 0A (f) c( ka ) (g) ( c k)( A B ) Solution (a) Adding the matrices A and B and multiplying the result by k = 5 we have k ( A+ B) = = 5 =

28 Chapter : Linear Equations and Matrices 8 (b) By heorem (.6) (b) we have k( ) A+ B = ka+ k B therefore by part (a) we have ka+ kb= k( A+ B) = (c) We first evaluate ( + and then scalar multiply this result by matrix B: c k) ( c+ k) B= ( + 5) B [ Substituting c= and k = 5] Multiplying each = B = = entry by c+ k B= cb+ kb therefore using our answer to part (d) By heorem (.6) (c) we have ( ) (c) we have 5 4 cb+ kb= ( c+ k) B= (e) How do we evaluate 0A? Multiply each entry of matrix A by 0 : A = 0 = (f) What is c( ka ) equal to? By heorem (.6) (a) we have ( ck ) = c( k ) ( ) = ( ck) = 0 A A and since ck = 5= 0 therefore c ka A A. he evaluation of 0A was done in part (e) above, therefore c( ka) = 0A = (g) How do we find c k? ( )( A B) First we evaluate c k and then subtract the matrices A B. o get the answer to c k A B) we multiply our results. Well ( c k) = 5= 7 and ( )( Hence we have A B= Subtracting the = corresponding entries 4 = Multiplying each = 77 7 entry by 7 ( c k)( A B) C Properties of Matrix Multiplication he algebraic rules of matrices are closely related to the algebraic rules of real numbers but there are some differences which you need to be aware of. Can you remember one difference between matrix and real number algebraic rules from the end of the last section (B)? Matrix multiplication is not commutative which means that the order we multiply matrices matters. If A and B are matrices then in general

29 Chapter : Linear Equations and Matrices 9 (.7) AB BA [ Not Equal] here may be cases where AB and BA are equal but generally they are not equal. In Example 8 from the last section (B) we were given the matrices A= and B = and we found that AB = but BA could not evaluated because the number of columns () of the Left Hand matrix, B, is not equal to the number of rows () of the ight Hand matrix, A. Also in Exercise (b) question parts (i) and (j) we were given B= 8 7 and D= Multiplication of these gives BD = 7 7 and DB = 4 0. What can we conclude about BD and DB? BD DB [Not Equal] You need to remember this important difference between matrix multiplication and multiplication of real numbers. We say matrix multiplication is not commutative. Proving properties of matrix multiplication can be a tedious task when examining all the entries. herefore we will only prove a particular property for one arbitrary entry and conclude that it is valid for all entries. For example consider the following matrix multiplication AB: jth Column ( AB) L ( AB) L( AB) a a L a j n r b L b j Lb n a ( ) ( ) ( ) a a r b j n b j b L L AB L AB L AB n M M M M M M M M M M M M = ith ow ai ai Lair ( AB) L ( AB) L( AB) i ij in br brj b M M M M L rn M M M M am am a L mr ( AB) L ( AB) L( AB) m mj mn Multiplying out the boxed (ith) row and (jth) column gives: ab + ab + L + ab = AB ( ) i j i j ir rj ij We will generally show the matrix multiplication results for the ith row and jth column. he entry in the ith row and jth column of a matrix AB is normally denoted by ( AB. Hence in this case we have ( AB) ow ( A) Column ( B ) = ij i where i =,,, L, m and j =,,, L, n. Writing out the complete row and column gives j ) ij

30 Chapter : Linear Equations and Matrices 0 (.8) ( AB) = ( a a a L a ) ij i i i ir b j b j M b rj ( ) ( ) ( ) L ( ) = ab + ab + ab + + ab = a b i j i j i j ir rj ik Kj K = r ik Kj ik Kj K = he sigma notation a b means summing the term a b from K = to K r = r. his sigma notation is difficult to understand but this compact way of writing a sum is normal in mathematics and you will see this in most mathematical literature. Note that ( AB) ij is the ij entry of matrix AB. We can evaluate certain entries of the matrix multiplication AB by using this formula as the next example illustrates. Example Let A= 4 7 and B= AB, AB, AB, AB and AB). Determine the following elements ( ) ( ) ( ) ( ) ( Solution What does the notation ( AB) mean? 4 4 ( AB ) means st row of matrix A times the nd column of matrix B. We have ( AB ) = ( 6) = ( ) + ( ) + ( 6 4) 0 = 4 What does the notation ( AB) mean? ( AB ) means nd row of matrix A times the rd column of matrix B. We have Similarly we have ( ) 5 AB = = ( 4 5 ) + ( 7 ( 7) ) + ( 9) = 78 9 AB is the rd row of matrix A times the st column of matrix B. ( ) ( ) ( ) ( AB ) = ( 5 ) = ( 5 ( ) ) + ( ) + ( 6) = 6 What does the notation ( AB) 4 mean? ( AB ) 4 is the 4 th row of matrix A times the rd column of matrix B but matrix A only has rows so we cannot evaluate ( AB) 4. ( AB ) 4 is the rd row of matrix A times the 4 th column of matrix B

31 Chapter : Linear Equations and Matrices 6 ( AB ) = ( 5 ) = ( 5 ( 6) ) + ( ) + ( ) 4 = Hence in summary we have ( AB ) = 0, ( AB ) = 78, ( ) = AB but we cannot evaluate ( AB) 4 because matrix A does not have 4 rows. AB, ( ) = 4 Note that we can work out ( AB ) 4 = but not ( AB ) 4. Be careful in evaluating ( AB ) ij because the orders of the row i and column j matters, that is for a general matrix C we have C C [Not Equal]. ij ji Here are some more results of matrix multiplication. heorem (.9). Properties of Matrix Multiplication. Let A, B and C be matrices of an appropriate size so that the following arithmetic operation can be carried out. hen (a) AB C = A BC) (Associative law for multiplication). (b) (c) ( ) ( ( + ) = + A B C AB AC(Left Distributive Law). ( B+ C) A= BA+ CA (ight Distributive Law). (d) A O= O A= O. Note that multiplication is commutative in this case. We only prove part (b). You are asked to prove the remaining parts in the Exercise (c). Proof of part (b) is not a difficult proof but it can be challenging to follow because of the notation involved and it is easy to confuse the i s and j s. If you get stuck do not move on but try to see if you understand the particular detail. Proof of (b). Let A be a matrix of size m r(m rows by r columns), B and C be matrices of size r n (r rows by n columns). Also let the entries in matrices A, B and C be denoted by a, b and c jk respectively. [he subscripts ij and jk gives the position of the entry]. We prove this result by considering an arbitrary row and column. Need to prove A B+ C = AB+ AC. First we look at the Left Hand Side of this equation ( ) and then we examine the ight and show they are equal. Adding the two matrices inside the brackets on the Left Hand Side gives: b + c b + cl b j + c j Lb n + c n b + c b + c L b j + c j Lb n + cn B+ C= M M M M br+ cr br + cr brj + crj brn + c L L rn Writing out the Left Hand Side A( B+ C ) we have a a L a r jth Col a a ar b + c b + c L bj + cj Lbn + c n M M M M b + c b + c b L j + c j Lbn + cn ith ow ai ai Lair M M M M b b r + cr br + cr rj + crj brn + crn M M M M L L am am a L mr ij jk

32 Chapter : Linear Equations and Matrices A B+ C we have Hence expanding for the ith row and jth column entry of ( ) ij ( + ) = ai ( bj + cj) + ai( b j + c j) + + air( brj + crj) A B C L ( ) ij Now we examine the ight Hand Side, AB + AC. By using the above formula (.8) we have for the ith row and jth column entry ( AB) = ab i j + ab ij i j + L + ab ir rj Similarly we have ( AC) = ac i j + ac ij i j + L + ac ir rj Adding these gives AB + AC = ab + ac + ab + ac + L+ ab + ac ( ) ( ) ij ij i j i j i j i j ir rj ir rj ( ) ( ) L ( ) = ai bj + cj + ai b j + c j + + air brj + crj ( C) By ( ) = A B+ Since we have ( ) + ( ) = ( + ) ij ij ij ij Factorizing out the a's AB AC A B C for arbitrary row and column entry therefore we have our general result, AB + AC = A( B + C ). Example Let A=, B= and C= 5. Evaluate (a) ( AB) C (b) ABC ( ) (c) A( B+ C ) (d) AB + AC (e) ( B+ C) A (f) BA + CA (g) AO (h) BO (i) CO (j) OC where O is the zero matrix of the appropriate size so that matrix multiplication can be evaluated. Solution (a) Multiplying the bracket term first: ( AB) = = 5 78 ( AB) C = 5 5 = 5 9 AB C = A BC therefore (b) By heorem (.9) (a) we have ( ) ( ) = = 78 AB C 5 ( ) ( ) A BC [By part (a)]

33 Chapter : Linear Equations and Matrices (c) Since matrices B and C are of different size therefore we cannot add them. hat is we cannot evaluate + which means A B+ C is impossible. B C ( ) (d) By heorem (.9) (b) we have AB + AC = A( B + C ) therefore we also cannot evaluate AB + AC because we cannot add B and C. (e) As part (c) above we cannot add matrices B and C. Hence we cannot evaluate B+ C A. ( ) (f) By heorem (.9) (c) we have BA + CA = ( B + C) A therefore we cannot evaluate BA + CA. For the remaining parts we use heorem (.9) (d) which says that A O= O A= O For parts (g) and (h) we have AO = =, BO = = For part (i) we need to more careful to select the correct size of the zero matrix. What is the O matrix equal to for the given matrix C? Since C= 5 therefore O = ( 0 0 0). emember for matrix multiplication, CO, the 9 number of columns of the Left Hand matrix, C, must equal the number of rows of the ight Hand matrix, O. We have CO = 5 ( 0 0 0) = ( 0) = 0 9 (j) By heorem (.9) (d) we have OC = CO = { 0 By Part (i) C4 Matrix Powers We define the matrix powers for square matrices only. What is a square matrix? A square matrix has an equal number of rows and columns so any n n (n by n) matrix is a square matrix. Let A be a square matrix then we define k (.0) A = A4 A4444 A L A k copies k Why can t we define A for non square matrices? Suppose matrix A is m n (m rows by n columns) with entries denoted by a and m does ij not equal n. What size matrix do we get if we evaluate A? a L a n a L a n A = M M M M M M am a mn am a L L mn

34 Chapter : Linear Equations and Matrices 4 Well we cannot carry out this matrix multiplication because the number of columns of the Left Hand matrix is n but the number of rows of the ight Hand matrix is m and since these are not equal therefore we cannot multiply. If we cannot find A then we cannot evaluate k A for k. Hence formula (.0) is only valid if A is a square matrix. Example Let A = and B=. Determine 8 (a) A (b) B (c) AB (d) ( ) (e) A+ B (f) Why is the following result not valid ( ) (g) What does ( A+ B equal? ) A+ B = A + AB+ B? Solution (a) emember A = A A so we have A = = (b) Similarly we have B = = 6 7 (c) How do we work out AB? Multiply the matrices A and B and then multiply our result by : AB = 8 6 = = 4 66 ( ) ( ) ( ) ( (d) How do we find A+ B? emember the result: (e) o find A + AB+ B A+ B = A+ B A+ B). First we add the matrices A and B and then square A+ B= + = 8 ( ) A+ B = = 9 69 A + AB+ B we add the above parts (a), (b) and (c): A + AB+ B = = = A AB = B = = Note that parts (d) and (e) are not equal. hat is Why not? ( A+ B) A + AB+ B [ Not Equal]

35 Chapter : Linear Equations and Matrices 5 (f) Because matrix multiplication is not commutative that means order of multiplication matters. emember the matrix multiplication AB does not equal BA. What is A+ B) equal to? ( ( A+ B) = ( A+ B) ( A+ B) = A + AB+ BA+ B Since AB BA [Not Equal] therefore AB + BA AB [Not Equal]. Let s check the result for this particular example. We know what AB is by part (c) above but what is BA equal to? BA = = 8 8 (g) Next we work out A + AB+ BA+ B. By the above parts we have A + AB+ BA+ B = = = = A AB BA = B = = 9 69 his is the same answer to part (d). herefore we have ( ) A+ B = A + AB+ BA+ B in this case. As seen in the last example you need to be very careful in carrying over our real number algebra to matrix algebra. It does not always work. We have the general rule: ( ) A+ B = A + AB+ BA+ B provided A and B are square matrices. SUMMAY Generally the rules of real number algebra can be applied to matrix algebra but there are some serious differences. For example if A and B are matrices then AB BA Not Equal [ ] We cannot blindly apply the properties of real number algebra to matrix algebra. Some properties of matrix addition are as follows: (.5) (a) A+ B = B+ A (b) ( A+ B) + C= A+ ( B+ C ) (c) A+ A+ A+ L+ A =na We can find particular entries of matrix multiplication AB by using (.8) ( AB) = ( ab i j) + ( ab i j) + ( ab ij i j) + L + ( ab ir rj) Matrix multiplication properties are: (.9) (a) ( AB) C = A BC) (Associative law for multiplication). ( ) ( (.9) (b) A B+ C = AB+ AC(Left Distributive Law). (.9) (c) ( B+ C) A= BA+ CA (ight Distributive Law). Let A and B be a square matrices then k (.0) A = A4 A4444 A L A k copies n copies