MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a 2 x + a 22 x 2 + + a 2n x n b 2 a 3 x + a 32 x 2 + + a 3n x n b 3 + + + () a m x + a m2 x 2 + + a mn x n b m In this system, a ij is a given real number as is b i Specifically, a ij is the coefficient for the unknown x j in the i th equation We call the set of all the a ij arranged in a rectangular array the coefficient matrix of the system Using matrix notation we can write the system as Ax b a a 2 a n a 2 a 22 a 2n x b a 3 a 32 a 3n x 2 b 2 x a m a m2 a n b m mn We define the augmented coefficient matrix for the system as  Consider the following matrix A and vector b We can then write A a a 2 a n b a 2 a 22 a 2n b 2 a 3 a 32 a 3n b 3 a m a m2 a mn b m 2 2 5 2, b 3 4 2 (2) (3) 3 8 (4) 4 Date: 26 March 2008

2 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS for the linear equation system 2 2 5 2 3 4 2 Ax b x x 2 3 8 4 x 3 (5) x + 2x 2 + x 3 3 2x + 5x 2 + 2x 3 8 3x 4x 2 2x 3 4 (6) 2 Row-echelon form of a matrix 2 Leading zeroes in the row of a matrix A row of a matrix is said to have k leading zeroes if the first k elements of the row are all zeroes and the (k+) st element of the row is not zero 22 Row echelon form of a matrix A matrix is in row echelon form if each row has more leading zeroes than the row preceding it 23 Examples of row echelon matrices The following matrices are all in row echelon form A 3 4 7 0 5 2 0 0 4 B C 0 0 0 2 0 0 0 3 0 4 0 0 3 (7) The matrix C would imply the equation system x + 3x 2 + x 3 b 4x 2 + x 3 b 2 3x 3 b 3 which could be easily solved for the various variable after dividing the third equation by 3 24 Pivots The first non-zero element in each row of a matrix in row-echelon form is called a pivot For the matrix A above the pivots are 3,5,4 For the matrix B they are,2 and for C they are,4,3 For the matrix B there is no pivot in the last row

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 3 25 Reduced row echelon form A row echelon matrix in which each pivot is a and in which each column containing a pivot contains no other nonzero entries, is said to be in reduced row echelon form This implies that columns containing pivots are columns of an identity matrix In the matrix D in equation (8) all three columns are pivot columns In the matrix E in equation (8) all the first two columns are pivot columns The matrices D and E are both in reduced row echelon form D 0 0 0 0 0 0 E 0 0 (8) 0 0 0 0 0 In matrix F in equation (9), columns 2, 4 and 5 are pivot columns The matrix F is in row echelon form but not in reduced row echelon form because of the in the a 25 and the 3 in the a 5 elements of the matrix The 5 in the a 3 element is irrelevant because column 3 is not a pivot column 0 5 0 3 F 0 0 0 0 0 0 0 (9) 0 0 0 0 0 26 Reducing a matrix to row echelon or reduced row echelon form Any system of equations can be reduced to a system with the A portion of the augmented matrix à in echelon or reduced echelon form by performing what are called elementary row operations on the matrix à These operations are the same ones that we used when solving a linear system using the method of Gaussian elimination There are three types of elementary row operations Performing these operations does not change the basic nature of the system or its solution : Changing the order in which the equations or rows are listed produces an equivalent system 2: Multiplying both sides of a single equation of the system (or a single row of Ã) by a nonzero number (leaving the other equations unchanged) results in a system of equations that is equivalent to the original system 3: Adding a multiple of one equation or row to another equation or row (leaving the other equations unchanged) results in a system of equations that is equivalent to the original system 27 Rank Definition (Rank) The number of non-zero rows in the row echelon form of an m n matrix A produced by elementary operations on A is called the rank of A Alternatively, the rank of an m n matrix A is the number of pivots we obtain by reducing a matrix to row echelon form Matrix D in equation 8 has rank 3, matrix E has rank 2, while matrix F in 9 has rank 3 Definition 2 (Full column rank) We say that an m n matrix A has full column rank if r n If r n, all columns of A are pivot columns The matrix L is of full column rank L 0 0 0 0

4 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Definition 3 (Full row rank) We say that an m n matrix A has full row rank if r m If r m, all rows of A have pivots and the reduced row echelon form has no zero rows The matrix M is of full row rank M 0 0 0 0 0 4 0 0 7 28 Solutions to equations (stated without proof) a A system of linear equations with coefficient matrix A which is m n, a right hand side vector b which is m, and augmented matrix  has a solution if and only if rank (A) rank (Â) This means that we can write the b vector as a linear combination of the columns of A b A linear system of equations must have either no solution, one solution, or infinitely many solutions c If the rank of A r n < m, the linear system Ax b has no solution or exactly one solution In other words, if the matrix A is of full column rank and has less columns than rows, the system will be inconsistent, or will have exactly one solution d If the rank of A r m < n, the linear system Ax b will always have a solution and there will be an infinite number of solutions In other words, if the matrix A is of full row rank and has more columns than rows, the system will always have an infinite number of solutions e If the rank of A r where r< m and r < n, the linear system Ax b will either have no solutions or there will be an infinite number of solutions In other words, if the matrix A has neither full row nor column rank, the system will be inconsistent or will have an infinite number of solutions f If the rank of A r m n, the linear system Ax b will exactly one solution In other words, if the matrix A is square and has full rank, the system will have a unique solution A coefficient matrix is said to be nonsingular, that is, the corresponding linear system has one and only one solution for every choice of right hand side b, b 2,, b m, if and only if number of rows of A number of columns of A rank (A) 3 Solving systems of linear equations by finding the reduced echelon form of a matrix and back substitution To solve a system of linear equations represented by a matrix equation, we first add the right hand side vector to the coefficient matrix to form the augmented coefficient matrix We then perform elementary row and column operations on the augmented coefficient matrix until it is in row echelon form or reduced row echelon form If the matrix is in row echelon we can solve it by back substitution, if it is in reduced row echelon form we can read the coefficients off directly from the matrix Consider matrix equation 5 with its augmented matrix à 2 2 5 2 3 4 2 Ax b x x 2 3 8 x 3 4 à 2 3 2 5 2 8 3 4 2 4 (5)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 5 Step Consider the first row and the first column of the matrix Element a is a non-zero number and so is appropriate for the first row of the row echelon form If element a were zero, we would exchange row with a row of the matrix with a non-zero element in column Step 2 Use row operations to knock out a 2 by turning it into a zero This would then make the second row have one more leading zero than the first row We can turn the a 2 element into a zero by adding negative 2 times the first row to the second row 2 ( 2 3 ) ( 2 4 2 6 ) 2 4 2 6 2 5 2 8 0 0 2 This will give a new matrix on which to operate Call it à à 2 3 0 0 2 (0) 3 4 2 4 We can obtain à by premultiplying à by what is called the a 2 elimination matrix denoted by E 2 where E 2 is given by E 2 0 0 2 0 () 0 0 The first and third rows of E 2 are just rows of the identity matrix because the first and third rows of à and à are the same The second row of à is a linear combination of the rows of à where we add negative 2 times the first row of à to times the second row of à To convince yourself of this fact remember what it means to premultiply a matrix by a row vector In the following example premultiplying the matrix 2 2 5 3 4 by the vector ( 0 2 ) adds one times the first row to twice the third row as follows ( ) 0 2 2 2 5 ( 5 6 ) 3 4 For the case at hand we have à 0 0 2 0 0 0 2 3 2 5 2 8 2 3 0 0 2 (2) 3 4 2 4 3 4 2 4 Step 3 Use row operations to knock out a 3 by turning it into a zero This would then make the third row have one more leading zero than the first row We can turn the a 3 element into a zero by adding 3 times the first row to the third row

6 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 3 ( 2 3 ) ( 3 6 3 9 ) 3 6 3 9 3 4 2 4 0 2 5 This will give a new matrix on which to operate Call it Ã2 à 2 2 3 0 0 2 (3) 0 2 5 We can obtain Ã2 by premultiplying à by what is called the a 3 elimination matrix denoted by E 3 where E 3 is given by E 3 0 0 0 0 (4) 3 0 The first and second rows of E 3 are just rows of the identity matrix because the first and second rows of à and Ã2 are the same The third row of Ã2 is a linear combination of the rows of à where we add 3 times the first row of à to times the third row of à For the case at hand we have à 2 0 0 0 0 3 0 2 3 0 0 2 2 3 0 0 2 (5) 3 4 2 4 0 2 5 Step 4 Use row operations to knock out a 32 by turning it into a zero This would then make the third row have one more leading zero than the second row We can turn the a 32 element into a zero by adding negative 2 times the second row of Ã2 to the third row of Ã2 2 ( 0 0 2 ) ( 0 2 0 4 ) 0 2 0 4 0 2 5 0 0 This will give a new matrix on which to operate Call it Ã3 à 3 2 3 0 0 2 (6) 0 0 We can obtain Ã3 by premultiplying Ã3 by what is called the a 32 elimination matrix denoted by E 32 where E 32 is given by

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 7 E 32 0 0 0 0 (7) 0 2 The first and second rows of E 32 are just rows of the identity matrix because the first and second rows of à 2 and Ã3 are the same The third row of Ã3 is a linear combination of the rows of Ã2 where we add negative 2 times the second row of Ã2 to times the third row of Ã2 For the case at hand we have à 3 0 0 0 0 0 2 2 3 0 0 2 2 3 0 0 2 (8) 0 2 5 0 0 We can then read off the solution for x 3 because the third equation says that x 3 The second equation says that x 2 2 We then solve for x by substituting in the first equation as follows x + 2x 2 + x 3 3 x + 2(2) + () 3 x + 4 + 3 x 2 (9) 4 Solving systems of linear equations by finding the reduced row echelon form of a matrix To solve a system of linear equations represented by a matrix equation, we first add the right hand side vector to the coefficient matrix to form the augmented coefficient matrix We then perform elementary row and column operations on the augmented coefficient matrix until it is in reduced row echelon form We can read the coefficients off directly from the matrix Consider the system of equations in section 3 2 2 5 2 3 4 2 Ax b x x 2 3 8 x 3 4 à 2 3 2 5 2 8 3 4 2 4 After performing suitable row operations we obtained the matrix Ã3 in equation 8 à 3 (5) 2 3 0 0 2 (20) 0 0 From this equation system we obtained the solutions by back substitution To find the reduced row echelon form we need to eliminate all non-zero entries in the pivot columns We begin with column 3 The a 23 element is already zero, so we can proceed to the a 3 element We do this in step 5

8 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Step 5 Use row operations to knock out a 3 by turning it into a zero We can turn the a 3 element into a zero by adding negative times the third row to the first row ( 0 0 ) ( 0 0 ) 0 0 2 3 2 0 2 This will give a new matrix on which to operate Call it Ã4 à 4 2 0 2 0 0 2 (2) 0 0 We can obtain Ã4 by premultiplying Ã3 by what is called the a 3 elimination matrix denoted by E 3 where E 3 is given by For the case at hand we have à 4 0 0 0 0 0 E 3 0 0 0 (22) 0 0 2 3 0 0 2 2 0 2 0 0 2 (23) 0 0 0 0 The third column of the matrix has a pivot in the third row and zeroes in the first and second rows Step 6 Use row operations to knock out a 2 by turning it into a zero We can turn the a 2 element into a zero by adding negative 2 times the second row to the first row 2 ( 0 0 2 ) ( 0 2 0 4 ) 0 2 0 4 2 0 2 0 0 2 This will give a new matrix on which to operate Call it Ã5 à 5 0 0 2 0 0 2 (24) 0 0 We can obtain Ã5 by premultiplying Ã4 by what is called the a 2 elimination matrix denoted by E 2 where E 2 is given by

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 9 E 2 2 0 0 0 (25) 0 0 For the case at hand we have à 5 2 0 0 0 2 0 2 0 0 2 0 0 2 0 0 2 (26) 0 0 0 0 0 0 It is obvious that x -2, x 2 2, and x 3 5 Examples 5 Example of a system with one solution Consider the system of equations x x 2 x 3 written in matrix form as 3x + 4x 2 + 2x 3 8 2x + x 2 4x 3 0 (27) The augmented matrix is of the form x 3 Ax b 3 4 2 x x 2 8 2 4 0 à After appropriate row and column operations we obtain (28) 3 4 2 8 (29) 2 4 0 0 0 2 0 0 2 (30) 0 0 3 The matrix A is of rank 3 because we have 3 non-zero rows in the reduced row echelon form We can also see that it is of rank 3 because we have 3 pivots The answers to the systems are {x 2, x 2 2, x 3 3} 52 Example of a system with three equations and two unknowns and no solution Consider the system of equations x x 2 written in matrix form as 3x + 4x 2 2 2x + x 2 6 (3)

0 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Ax b ( ) 3 4 x (32) 2 x 2 2 6 The augmented matrix is of the form à 3 4 2 (33) 2 6 After appropriate row and column operations we obtain 0 0 6 5 (34) 0 0 The matrix A is of rank 2 because we have 2 non-zero rows in the reduced row echelon form We can also see that it is of rank 2 because we have 2 pivots The third equation is inconsistent and so the system has no solution 53 Example of a system with three equations and two unknowns with exactly one solution Consider the system of equations x x 2 written in matrix form as 3x + 4x 2 2 2x + x 2 5 (35) Ax b ( ) 3 4 x (36) 2 x 2 2 5 The augmented matrix is of the form à 3 4 2 (37) 2 5 Adding 3 times the first equation to the second equation we obtain 0 (38) 2 5 Adding -2 times the first equation to the third equation we obtain 0 (39) 0 3 3

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Adding -3 times the second equation to the third equation we obtain 0 (40) 0 0 0 Adding the second equation to the first equation we obtain ( ) 0 2 0 (4) 0 0 0 The matrix A is of rank 2 because we have 2 non-zero rows in the reduced row echelon form We can also see that it is of rank 2 because we have 2 pivots The third equation implies that 0 0 and the system has one solution We can also see this when we realize that the third row of à {2,, 5} is times the first row of à plus 3 times the second row of à ( ) ( ) 3 ( 3 4 2 ) ( 9 2 6 ) 9 2 6 2 5 54 Example of a system with two equations and three unknowns and infinite solutions Consider the system of equations written in matrix form as x x 2 x 3 3x + 4x 2 + 2x 3 8 (42) x 3 Ax b ( ) x x 3 4 2 2 The augmented matrix is of the form ( ) à 3 4 2 8 After appropriate row and column operations we obtain ( ) 0 2 4 0 5 ( ) (43) 8 (44) (45)

2 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS The matrix A is of rank 2 because we have 2 non-zero rows in the reduced row echelon form We can also see that it is of rank 2 because we have 2 pivots We can pick an arbitrary number for x 3 and then find values for x and x 2 by solving each equation Therefore there are infinite solutions For example, let x 3 3 We then have the following system ( ) 0 2 x ( ) x 0 2 4 5 3 We then have x + ( 2)(3) 4 We then have x 4 + 6 2 x 2 + ( )(3) 5 x 2 5 + 3 2 Substituting in equation 43 we obtain ( ) 2 ( ) 2 3 4 2 8 3 Picking a different value for x 3 we would obtain different values for x and x 2 55 Example of a system with three equations and three unknowns and infinite solutions Consider the system of equations x x 2 x 3 written in matrix form as 3x + 4x 2 + 2x 3 8 2x + 3x 2 + x 3 7 (46) Ax b 3 4 2 x x 2 (47) 8 2 3 x 3 7 The augmented matrix is of the form à 3 4 2 8 (48) 2 3 7 After appropriate row and column operations we obtain 0 2 4 0 5 (49) 0 0 0 0

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 3 The matrix A is of rank 2 because we have 2 non-zero rows in the reduced row echelon form We can also see that it is of rank 2 because we have 2 pivots We can pick an arbitrary number for x 3 and then find values for x and x 2 by solving each equation Therefore there are infinite solutions 56 Example of a system with three equations and three unknowns and no solution Consider the system of equations x x 2 + 2x 3 5 written in matrix form as 2x + 3x 2 x 3 5 3x + 2x 2 + x 3 9 (50) Ax b 2 2 3 x x 2 5 (5) 5 3 2 x 3 9 The augmented matrix is of the form à 2 5 2 3 5 (52) 3 2 9 After appropriate row and column operations we obtain 0 4 0 (53) 0 0 0 The matrix A is of rank 2 because we have 2 non-zero rows in the reduced row echelon form We can also see that it is of rank 2 because we have 2 pivots The third equation is inconsistent and so the system has no solution 6 Systems of linear equations and determinants 6 Solving a general 2x2 equation system using elementary row operations Consider the following simple 2x2 system of linear equations where the A matrix is written in a completely general form a x + a 2 x 2 b (54) a 2 x + a 22 x 2 b 2 We can write this in matrix form as Ax b [ ] [ ] [ ] a a A 2 x b (55), x, b a 2 a 22 x 2 b 2 If we append the column vector b to the matrix A, we obtain the augmented matrix for the system This is written as [ ] a a à 2 b (56) a 2 a 22 b 2

4 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS We can perform row operations on this matrix to reduce it to reduced row echelon form We will do this in steps The first step is to divide each element of the first row by a This will make the first pivot a one This will give à Now multiply the first row by a 2 to yield and subtract it from the second row [ [ a 2 b ] a a a 2 a 22 b 2 a 2 a 2 a 2 a a 2 b a ] (57) (58) a 2 a 22 b 2 a 2 a 2 a 2 b a 2 a a (59) 0 a 22 a 2 a 2 a b 2 a 2 b a This will give a new matrix on which to operate à 2 [ 0 a 2 a a a 22 a 2 a 2 b a a b 2 a 2 b a a Now multiply the second row by the reciprocal of the element in the a 22 position, ie to obtain à 3 [ 0 a 2 b a a a b 2 a 2 b a a 22 a 2 a 2 ] ] (60) a a a 22 a 2 a 2 Now multiply the second row by a2 a and subtract it from the first row First multiply the second row by a2 a to yield [ 0 ] a 2 a 2 ( a b 2 a 2 b ) a a ( a a 22 a 2 a 2 ) Now subtract the expression in equation 62 from the first row of Ã3 to obtain the following row (6) (62) [ ] [ a 2 b a a 0 ] [ a 2 a 2 (a b 2 a 2 b ) a a (a a 22 a 2 a 2) 0 b a ] a2 (a b2 a2 b) a (a a 22 a 2 a 2) (63) Now replace the first row in Ã3 with the expression in equation 63 to obtain Ã4 à 4 [ 0 0 b a a2 ( a b2 a2 b ) a ( a a 22 a 2 a 2 ) a b 2 a 2 b a a 22 a 2 a 2 This can be simplified by putting the upper right hand term over a common denominator, and canceling like terms as follows ] (64)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 5 0 Ã 4 0 0 0 0 0 b a 2 a 22 b a a 2 a 2 a 2 a 2 b 2 + a a 2 a 2 b a 2 (a a 22 a 2 a 2 ) a b 2 a 2 b a a 22 a 2 a 2 b a 2 a 22 a 2 a 2 b 2 a 2 (a a 22 a 2 a 2 ) a b 2 a 2 b a a 22 a 2 a 2 b a 22 a 2 b 2 a a 22 a 2 a 2 a b 2 a 2 b a a 22 a 2 a 2 We can now read off the solutions for x and x 2 They are (65) x b a 22 a 2 b 2 a a 22 a 2 a 2 x 2 a b 2 a 2 b a a 22 a 2 a 2 Each of the fractions has a common denominator It is called the determinant of the matrix A Note that this part of the solution of the equation Axb does not depend on b; it will be the same no matter what right hand side is chosen That is why it is called the determinant of the matrix A and not of the system Ax b The determinant of an arbitrary 2x2 matrix A is given by (66) det(a) A a a 22 a 2 a 2 (67) The determinant of the 2x2 matrix is given by difference in the product of the elements [ ] along the 3 main diagonal and the reverse main diagonal The determinant of the matrix A is 2-6 2 2-4 If we look closely at the solutions in equation 66, we can see that we can also write the numerator of each expression as a determinant In particular x b a 22 a 2 b 2 a a 22 a 2 a 2 x 2 a b 2 a 2 b a a 22 a 2 a 2 b a 2 b 2 a 22 A a b a 2 b 2 A The matrix of which we compute the determinant in the numerator of the first expression is the matrix A, where the first column has been replaced by the b vector The matrix of which we compute the determinant in the numerator of the second expression is the matrix A where the second column has been replaced by the b vector This procedure for solving systems of equations is called Cramer s rule and will be discussed in more detail later For those interested, here is a simple example

6 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 62 An example problem with Cramer s rule Consider the system of equations Using Cramer s rule 3x + 5x 2 (68) 8x 3x 2 3 b a 2 b 2 a 22 x A a b a 2 b 2 x 2 A 5 3 3 3 5 8 3 3 8 3 3 5 8 3 ( 33) (65) ( 9) (40) (39) (88) ( 9) (40) 49 49 98 49 2 (69) 2 Determinants 2 Introduction Associated with every square matrix is a number called the determinant of the matrix The determinant provides a useful summary measurement of that matrix Of particular importance is whether the determinant of a matrix is zero As we saw in section 6, elements of the solution of the equation Ax b for the 2x2 case can be written in terms of the determinant of the matrix A For a square matrix A, the equation Ax 0 has a non-trivial solution (x 0) if and only if the determinant of A is zero 22 Definition and analytical computation 22 Definition of a determinant The determinant of an n n matrix A a ij, written A, is defined to be the number computed from the following sum where each element of the sum is the product of n elements: A (±)a i a 2j a nr, (70) the sum being taken over all n! permutations of the second subscripts A term is assigned a plus sign if (i,j,,r) is an even permutation of (,2,,n) and a minus sign if it is an odd permutation An even permutation is defined as making an even number of switches of the indices in (,2,,n), similarly for an odd permutation Thus each term in the summation is the product of n terms, one term from each column of the matrix Note that the first element of each pair does not change We always have a a 2 a 3 a n where it is only the second subscript that is changing When there are no switches in the indices from the natural order, the product will be a a 22 a 33 a nn or the product of the terms on the main diagonal When the only switch is between the subscripts and 2 of the first two terms we obtain the product a 2 a 2 a 33 a nn This product will be assigned a minus sign If we switch subscripts and 3 of the first and third terms we obtain the product a 3 a 22 a 3 a nn which also has a minus sign

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 7 222 Example As a first example consider the following 2x2 case A ( ) 2 2 5 (7) Here a, a 2 2, a 2 2, and a 22 5 There are two permutations of (,2) : (,2) is an even permutation with 0 switches of the indices (,2): + a a 22 2: (2,) is odd permutation with switch of the indices: - a 2 a 2 Notice that the permutations are over the second subscripts The determinant of A is then A +a a 22 a 2 a 2 5 2 2 (72) The 2 x 2 case is easy to remember because there are just 2 rows and 2 columns Each term in the summation has two elements and there are two terms The first term is the product of the diagonal elements and the second term is the product of the off-diagonal elements The first term has a plus sign because the second subscripts are and 2 while the second term has a negative sign because the second subscripts are 2 and (one permutation of and 2) 223 Example 2 Now consider another 2x2 example A ( ) 3 5 8 3 (73) Here a 3, a 2 5, a 2 8, and a 22-3 The determinant of A is given by A +a a 22 a 2 a 2 (3)( 3) (8)(5) 49 (74) 224 Example 3 Consider the following 3x3 case A 2 3 0 5 2 (75) 0 4 The indices are the numbers, 2 and 3 There are 6 (3!) permutations of (, 2, 3) They are as follows: (, 2, 3), (2,, 3), (, 3, 2), (3, 2, ), (2, 3, ) and (3,, 2) The three in bold come from switching one pair of indices in the natural order (, 2, 3) The final two come from making a switch in one of the bold sets Specifically : (, 2, 3) is even (0 switches): + a a 22 a 33 2: (2,, 3) is odd ( switches): - a 2 a 2 a 33 3: (, 3, 2) is odd ( switches): - a a 23 a 32 4: (3, 2, ) is odd ( switches): - a 3 a 22 a 3 5: (2, 3, ) is even (2 switches): + a 2 a 23 a 3 6: (3,, 2) is even (2 switches): + a 3 a 2 a 32 Notice that the permutations are over the second subscripts, that is, every term is of the form a a 2 a 3 with the second subscript varying The determinant of A is then

8 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS A + a a 22 a 33 a 2 a 2 a 33 a a 23 a 32 a 3 a 22 a 3 +a 2 a 23 a 3 (76) +a 3 a 2 a 32 20 0 0 5 + 4 + 0 9 The first subscripts in each term do not change but the second subscripts range over all permutations of the numbers, 2, 3,, n The first term is the product of the diagonal elements and has a plus sign because the second subscripts are (, 2, 3) The last term also has a plus sign because the second subscripts are (3,, 2) This permutation come from 2 switches [(, 2, 3) (3,2,) (3,,2)] A simple way to remember this formula for a 3x3 matrix is to use diagram in figure The elements labeled j are multiplied together and receive a plus sign in the summation The elements labeled 2j are multiplied together and receive a minus sign in the summation as in figure 2 Figure 3 x 3 Determinant Plus Signs 225 Example 4 Consider the following 3x3 case A 2 4 2 3 (77) 2 0 The determinant of A is

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 9 Figure 2 3 x 3 Determinant Minus Signs A + a a 22 a 33 +a 2 a 23 a 3 +a 3 a 2 a 32 a 3 a 22 a 3 a 2 a 2 a 33 (78) a a 23, a 32 226 Example 5 Consider the following 4x4 case 0 + 6 + 6 ( 4) 0 6 20 a a 2 a 3 a 4 A a 2 a 22 a 23 a 24 a 3 a 32 a 33 a 34 (79) a 4 a 42 a 43 a 44 The indices are the numbers, 2, 3, and 4 There are 24 (4!) permutations of (, 2, 3, 4) They are contained in table Notice that the permutations are over the second subscripts, that is, every term is of the form a a 2 a 3 a 4 with the second subscript varying The general formula is rather difficult to remember or implement for large n and so another method based on cofactors is normally used 23 Submatrices and partitions Before discussing the computation of determinants using cofactors a few definitions concerning matrices and submatrices will be useful

20 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Table Determinant of a 4x4 Matrix Permutation Sign element of sum Permutation Sign element of sum 234 + a a 22 a 33 a 44 342 - a 3 a 2 a 34 a 42 243 - a a 22 a 34 a 43 324 + a 3 a 2 a 32 a 44 324 - a a 23 a 32 a 44 324 - a 3 a 22 a 3 a 44 342 + a a 23 a 34 a 42 324 + a 3 a 22 a 34 a 4 432 - a a 24 a 33 a 42 342 - a 3 a 24 a 32 a 4 423 + a a 24 a 32 a 43 342 + a 3 a 24 a 3 a 42 234 - a 2 a 2 a 33 a 44 423 - a 4 a 2 a 32 a 43 243 + a 2 a 2 a 34 a 43 432 + a 4 a 2 a 33 a 42 234 - a 2 a 23 a 34 a 4 423 - a 4 a 22 a 33 a 4 234 + a 2 a 23 a 3 a 44 423 + a 4 a 22 a 3 a 43 243 - a 2 a 24 a 3 a 43 432 - a 4 a 23 a 3 a 42 243 + a 2 a 24 a 33 a 4 432 + a 4 a 23 a 32 a 4 23 Submatrix A submatrix is a matrix formed from a matrix A by taking a subset consisting of j rows with column elements from a set k of the columns For example consider A({,3},{2,3}) below A 3 4 7 2 5 2 0 4 3 4 7 A({, 3}, {2, 3}) 2 5 2 0 4 ( ) 4 7 0 4 (80) The notation A({,3},{2,3}) means that we take the first and third rows of A and include the second and third elements of each row 232 Principal submatrix A principal submatrix is a matrix formed from a square matrix A by taking a subset consisting of n rows and column elements from the same numbered columns For example consider A({,3},{,3}) below

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 2 A 3 4 7 2 5 2 0 4 3 4 7 A({, 3}, {, 3}) 2 5 2 0 4 ( ) 3 7 4 (8) 233 Minor A minor is the determinant of a square submatrix of the matrix A For example consider A({2, 3}, {, 3}) A 3 4 7 2 5 2 0 4 3 4 7 A({2, 3}, {, 3}) 2 5 2 A({2, 3}, {, 3}) 6 0 4 ( ) 2 2 4 234 Principal minor A principal minor is the determinant of a principal submatrix of A For example consider A({, 2}, {, 2}) A 3 4 7 2 5 2 0 4 3 4 7 A({, 2}, {, 2}) 2 5 2 A({, 2}, {, 2}) 7 0 4 ( ) 3 4 2 5 235 The minor associated with the element a ij of a square matrix A We are often interested in the principle minor formed by including all rows of A except the i th and all columns except the j th This is formed by deleting the i th row and the j th column of A We use the notation M ij to denote this minor associated with the ij th element of A (82) (83)

22 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS A M 22 3 4 7 2 5 2 0 4 3 4 7 2 5 2 0 4 ( ) 3 7 4 (84) 236 Leading principal minor Let A (a ij ) be any n n The leading principle minors of A are the n determinants: a a 2 a k a 2 a 22 a 2k D k, k, 2,, n (85) a k a k2 a kk D k is obtained by crossing out the last n-k columns and n-k rows of the matrix Thus for k, 2, 3, n, the leading principle minors are, respectively a, a a 2 a 2 a 22, a a 2 a 3 a a 2 a n a 2 a 22 a 23 a 3 a 32 a 33,, a 2 a 22 a 2n (86) a n a n2 a nn 237 Cofactor The cofactor (denoted A ij ) of the element a ij of any square matrix A is (-) i+j times the minor of A that is obtained by including all but the i th row and the j th column, or alternatively the minor that is obtained by deleting the i th row and the j th column For example consider the matrix A A a a 2 a 3 a 2 a 22 a 23 3 4 7 2 5 2 (87) a 3 a 32 a 33 0 4 To find the cofactor of a 2 we first find the submatrix that includes all rows and columns except the first and second A({2, 3}, {, 3}) 3 4 7 ( ) 2 5 2 2 2 4 0 4 We then multiply the determinant of this matrix by (-) i+j For the example element a 2, i+j +2 3 and yields cofactor(a 2 ) A 2 ( ) 3 A({2, 3}, {, 3} ( ) 3 2 2 4 (89) ( ) 3 (6) 6 (88)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 23 24 Computing determinants using cofactors 24 Definition of a cofactor expansion The determinant of a square matrix A can be found inductively using the following formula det A A n j a ij A ij (90) where i denotes the i th row of the matrix A This is called an expansion of A by column i of A The result is the same for any other row This can also be done for columns letting the sum range over i instead of j 242 Examples : Consider as an example the following 2x2 matrix and expand using the first row ( ) 4 3 B 2 B 4 ( ) 2 (2) + 3 ( ) 3 () 8 3 5 where the cofactor of 4 is (-) (+) times the submatrix that remains when we delete row and column from the matrix B while the cofactor of 3 is (-) (+2) times the submatrix that remains when we delete row and column 2 from the matrix B In this case each the principle submatrices are just single numbers so there is no need to formally compute a determinant 2: Now consider a 3x3 matrix B B 2 3 0 5 2 0 4 If we expand using the first row of the matrix, the relevant minors are 2 3 cofactor(b ) B ({2, 3}, {2, 3}) 0 5 2 0 4 2 3 cofactor(b 2 ) B ({2, 3}, {, 3}) 0 5 2 0 4 2 3 cofactor(b 3 ) B ({2, 3}, {, 2}) 0 5 2 0 4 (9)

24 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS The determinant is then B ( ) 2 5 2 0 4 + 2 0 2 ( )3 4 + 3 0 5 ( )4 0 () (20) + ( 2) ( 2) + 3 ( 5) 9 We can also compute it using the third row of the matrix B 2 3 0 5 2 0 4 B ( ) 4 2 3 5 2 + 0 3 ( )5 0 2 + 4 2 ( )6 0 5 or the first column () ( ) + (0) (2) + (4) (5) 9 B 2 3 0 5 2 0 4 B ( ) 2 5 2 0 4 + 0 2 3 ( )3 0 4 + 2 3 ( )4 5 2 () (20) + (0) ( 8) + () ( ) 9 3: Consider as another example the following 3x3 matrix A 2 4 2 3 (92) 2 0 The determinant of A computed using the cofactor expansion along the third row is given by A ( ) 4 2 4 3 + 2 4 ( )5 2 3 + 0 2 ( )6 2 ( ) ( 0) + ( 2) ( 5) + (0) ( 5) 20 25 Expansion by alien cofactors 25 Definition of expansion by alien cofactors When we expand a row of the matrix using the cofactors of a different row we call it expansion by alien cofactors a kj A ij 0 k i (93) j

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 25 252 Some examples : example 4 3 2 Expand row using the 2nd row cofactors to obtain (4) (-) (2+) (3) + (3) (-) (2+2 (4) -2 + 2 0 2: example 2 4 3 6 2 3 3 Expand row using the 2nd row cofactors The second row cofactors are ( ) (2+) 3 3 and ( ) (2+2) 4 and ( ) (2+3) 4 3 3 This then gives for the cofactor expansion(4) (-) (0) + (3) () (3) + () (-) (9) 0 253 General principle If we expand the rows of a matrix by alien cofactors, the expansion will equal zero 26 Singular and nonsingular matrices The square matrix A is said to be singular if A 0, and nonsingular if A 0 26 Determinants, Minors, and Rank Theorem The rank of an n n matrix A is k if and only if every minor in A of order k + vanishes, while there is at least one minor of order k which does not vanish Proposition Consider an n n matrix A : det A 0 if every minor of order n - vanishes 2: If every minor of order n equals zero, then the same holds for the minors of higher order 3 (restatement of theorem): The largest among the orders of the non-zero minors generated by a matrix is the rank of the matrix

26 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 27 Basic rules for determinants : The determinant of an n n identity matrix is [ ] 0 0, 0 0 0 0 0 0 0 0 0 2: If two rows A are interchanged, then the determinant of A changes sign, but keeps its absolute value If two columns of A are interchanged, then the determinant of A changes sign, but keeps its absolute value 3: The determinant is a linear function of each row separately If the first row is multiplied by α, the determinant is multiplied by α If the first rows of two matrices are added, the determinants are added This rule applies only when other rows do not change Consider the following example matrices α a α a 2 α a 3 α a n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn a + b a 2 + b 2 a 3 + b 3 a n + b n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn a a 2 a 3 a n a 2 a 22 a 23 a 2n α a n a n2 a n3 a nn a a 2 a 3 a n a 2 a 22 a 23 a 2n + a n a n2 a n3 a nn b b 2 b 3 b n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn Or consider the following numerical example 3 5 0 2 7 0 0 4 0 0 0 2 7 0 0 4 + 0 3 5 0 2 7, (First row of matrix on left is sum of two on right) 0 0 4 0 0 0 2 0 0 0 4 + 0 0 0 0 7 0 0 4 + 0 3 5 0 2 7 0 0 4, Second rows of first two matrices sum to second rows of matrices above 0 0 2 0 0 0 0 4 + 0 0 0 0 7 0 0 4 + 0 3 5 0 2 7, (Factor 2 out of second row of first matrix above) 0 0 4 0 0 (2)(4) 0 0 0 0 + 0 0 0 0 7 0 0 4 + 0 3 5 0 2 7, (Factor 4 out of third row of first matrix above) 0 0 4 0 0 8 + 0 0 7 0 0 4 + 0 3 5 0 2 7 0 0 4 (94)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 27 Property 3 can be used to show that if A is an n n matrix and α is a real number then αa α n A 4: If two rows or columns of A are equal, then A 0 5: If a scalar multiple of one row (or column) of A is added to another row (or column) of A, then the value of the determinant does not change This property can be used to show that if two rows or columns of A are proportional, then A 0 If we multiply one of the rows by the factor of proportionality and add it to the other row, we will obtain a row of zeroes and a row of zeroes implies det A 0 6: A matrix with a row (or column) of zeros has A 0 Given this property of determinants, the determinant in equation 94 is equal to 8 as the determinants of the two matrices on the right are both zero 7: If A is triangular, then det A A a a 22 a nn the product of the diagonal entries The product of the diagonal entries in equation 94 is equal to 8 This property, of course, implies that the determinant of a diagonal matrix is the product of the diagonal elements 8: If A is singular (to be defined later), then the determinant of A 0 If A is invertible (to be defined later), then det A 0 9: If A and B are both n n then AB A B 0: The determinant of A is equal to the determinant of A, ie, A A : The determinant of a sum is not necessarily the sum of the determinants 28 Student problem Provide an example for each of items - in section 27

28 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 3 The inverse of a matrix A 3 Definition of the inverse of a matrix Given a matrix A, if there exists a matrix B such that AB BA I (95) then we say that B is the inverse of A Moreover, A is said to be invertible in this case Because BA AB I, the matrix A is also an inverse of B that is, A and B are inverses of each other Inverses are only defined for square matrices We usually denote the inverse of A by A and write A A AA I (96) 32 Some examples of matrix inversion Show that the following pairs of matrices are inverses of each other [ 2 ] [ 2 3 3 3 3 ] 4 3 6 2 3 3 3 0 3 7 7 6 3 2 7 2 7 0 2 33 Existence of the inverse A square matrix A has an inverse A 0 A square matrix is said to be singular if A 0 and nonsingular if A 0 Thus a matrix has an inverse if and only if it is nonsingular 34 Uniqueness of the inverse The inverse of a square matrix is unique if it exists 35 Some implications of inverse matrices AX I X A (97) BA I B A 36 Properties of the inverse Let A and B be invertible n n matrices : A is invertible and (A ) A 2: AB is invertible and (AB) B A 3: A is invertible and (A ) (A ) 4: (ABC) C B A, if A, B, and C have inverses 5: (ca) c A whenever c is a number 0 6: In general, (A + B) A + B 37 Orthogonal matrices A matrix A is called orthogonal if its inverse is equal to its transpose, that is if A A (98)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 29 38 Finding the inverse of a 2x2 matrix For a general 2x2 matrix A, we can find the inverse using the following matrix equation We can divide this into two parts as follows AB I (99) [ ] [ ] [ ] a a 2 b b 2 0 a 2 a 22 b 2 b 22 0 [ ] [ ] [ ] a a 2 b a 22 b 2 0 a 2 [ a a 2 a 2 a 22 ] [ b2 b 22 ] and [ ] 0 We can solve each system using Cramer s rule For the first system we obtain b a 2 0 a 22 a a 2 a 2 a 22 and a a 2 0 b 2 a a 2 a 2 a 22 In a similar fashion we can show that a 22 a a 22 a 2 a 2 a 2 a a 22 a 2 a 2 b 2 a 2 a a 22 a 2 a 2 (00) Combining the expressions we see that A a b 22 a a 22 a 2 a 2 a a 22 a 2 a 2 [ ] a22 a 2 a 2 a (0) This leads to the familiar rule that to compute the determinant of a 2x2 matrix we switch the diagonal elements, change the sign of the off-diagonal ones, and divide by the determinant 39 Finding the inverse of a general n n matrix The most efficient way to find the inverse of an n n matrix is to use elementary row operations or elimination on the augmented matrix [ A : I ] Or in other words solve the following systems of equations for b, b 2,, b n

30 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS a a 2 a 3 a n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn b b 2 b n 0 0 a a 2 a 3 a n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn b 2 b 22 b n2 0 0 a a 2 a 3 a n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn b n b 2n b nn 0 0 (02) As discussed in section 4, this is done by performing elimination on the matrix a a 2 a 3 a n 0 0 0 a 2 a 22 a 23 a 2n 0 0 0 a n a n2 a n3 a nn 0 0 0 (03) 4 Finding the inverse of a matrix using elementary row operations The most effective computational way to find the inverse of a matrix is to use elementary row operations This is done by forming the augmented matrix [A : I] similar to the augmented matrix we use when solving a system equation, except that we now add n columns of the identity matrix instead of the right hand side vector in the equations system In effect we are writing the system A X I (04) and then solving for the matrix X, which will be A 4 Example with 2x2 matrix Consider the following matrix A ( 2 2 3 ) (05) Consider solving the following equation Ax b ( 2 2 3 ) ( x x 2 ) ( 0) (06) We can solve the system by appending the right hand side vector, b, to the matrix A to form the matrix à and then use elimination to create an identity matrix in place of the A matrix We do this as follows

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 3 Ã ( ) 2 2 3 0 Add negative 2 times the first row to the second row to obtain Now divide the second row by - to obtain 2 4 2 2 3 0 0 2 ( ) 2 Ã 0 2 Ã 2 ( 2 ) 0 2 Now multiply the second row by -2 and add to the first row (07) (08) (09) 0 2 4 2 0 3 ( ) 0 3 Ã3 0 2 We can see that x -3 and x 2 2 So if we post multiply A by column of the identity matrix If we instead solve the system ( ) ( ) 2 3 2 3 2 ( 0) ( 0 Ax ) (0) ( ) 3 we obtain the first 2 () ( ) ( ) ( (2) 2 x 0 2 3 x 2 ) we obtain a vector that when premultiplied by the matrix A gives us the second column of the identity matrix So we can find the inverse in general by solving the system x x 2 x n 0 0 0 x 2 x 22 x 2n 0 0 0 A x 3 x 32 x 3n 0 0 0 x n x n2 x n 0 0 0 So for our example append an identity matrix as follows (3)

32 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS à ( 2 ) 0 2 3 0 Add negative 2 times the first row to the swecond row to obtain Now divide the second row by - to obtain 2 4 2 0 2 3 0 0 2 ( ) 2 0 à 0 2 à 2 ( 2 ) 0 0 2 Now multiply the second row by -2 and add to the first row The inverse of A is then given by We can check this as follows 0 2 4 2 2 0 0 3 2 ( ) 0 3 2 Ã3 0 2 A ( ) 3 2 2 ( ) ( ) 3 2 2 2 2 3 ( ) 0 0 (4) (5) (6) (7) (8) (9) 42 Example with 3x3 matrix Consider the following matrix A 2 2 0 4 (20) 0 2 Now append an identity matrix as follows à 2 2 0 0 0 4 0 0 (2) 0 2 0 0 Add negative times the first row to the last row to obtain

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 33 2 2 0 0 0 2 0 0 0 2 0 0 (22) à 0 2 4 2 0 0 0 0 0 2 0 0 Now divide the second row by 4 to obtain à 2 2 2 0 0 0 /4 0 /4 0 (23) 0 2 0 0 Now multiply the second row by 2 and add to the third row 0 2 /2 0 /2 0 0 2 0 0 0 0 /2 /2 (24) Ã3 2 2 0 0 0 /4 0 /4 0 0 0 /2 /2 Now multiply the third row by 2 à 4 2 2 0 0 0 /4 0 /4 0 (25) 0 0 2 2 Now multiply the third row by negative /4 and add to the second row 0 /4 0 /4 0 0 0 /4 /2 /4 /2 0 0 /2 0 /2 Ã5 2 2 0 0 0 0 /2 0 /2 0 0 2 2 Now multiply the third row by negative 2 and add to the first row 2 2 0 0 0 0 2 4 2 4 2 0 5 2 4 2 0 5 2 4 Ã6 0 0 /2 0 /2 0 0 2 2 (26) (27)

34 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Now multiply the second row by negative 2 and add to the first row The inverse of A is then given by We can check this as follows 2 0 5 2 4 0 2 0 0 0 0 4 2 3 0 0 4 2 3 Ã7 0 0 /2 0 /2 0 0 2 2 (28) 4 2 3 A /2 0 /2 (29) 2 2 2 2 4 2 3 0 0 0 4 /2 0 /2 0 0 (30) 0 2 2 2 0 0 5 Finding the inverse of a matrix and solving a system of equations simultaneously using elementary row operations We can, of course, find the inverse of a matrix and solve a system of equations simultaneously by creating a special augmented matrix that contains both the identity matrix and the b vector, ie à [A : I : b] and then solving for the matrix X Consider the following matrix and right hand side vector A X [I : b] (3) A 3 2 2 5 2 (32) 4 7 b 4 5 (33) 5 Now append an identity matrix and the b vector as follows à 3 2 0 0 4 2 5 2 0 0 5 (34) 4 7 0 0 5 Add 2 times the first row to the second row to obtain

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 35 Now divide the second row by - to obtain 2 6 4 2 0 0 8 2 5 2 0 0 5 0 2 2 0 3 Ã2 3 2 0 0 4 0 2 2 0 3 4 7 0 0 5 (35) Ã 22 3 2 0 0 4 0 2 2 0 3 (36) 4 7 0 0 5 Now multiply the first row by -4 and add to the third row 4 2 8 4 0 0 6 4 7 0 0 5 0 4 0 Ã3 3 2 0 0 4 0 2 2 0 3 0 4 0 Now multiply the second row by - and add to the third row 0 2 2 0 3 0 4 0 0 0 2 2 Ã32 3 2 0 0 4 0 2 2 0 3 0 0 2 2 Now multiply the third row by 2 and add to the second row 0 0 2 4 2 2 4 0 2 2 0 3 0 0 6 2 Ã23 3 2 0 0 4 0 0 6 2 0 0 2 2 Now multiply the third row by negative 2 and add to the first row (37) (38) (39)

36 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 0 0 2 4 2 2 4 3 2 0 0 4 3 0 5 2 2 0 3 0 5 2 2 0 Ã3 0 0 6 2 0 0 2 2 Now multiply the second row by 3 and add to the first row (40) The inverse of A is then given by The solution to the equation is given by 0 3 0 8 3 6 3 3 0 5 2 2 0 0 0 3 4 3 Ã2 0 0 3 4 3 0 0 6 2 0 0 2 2 (4) A 3 4 6 2 (42) 2 We can check this as follows We can also see that x 3 (43) 2 3 2 2 5 2 3 4 5 (44) 4 7 2 5 x A b 3 3 4 6 2 2 2 4 5 5 (45) 6 Solving equations by matrix inversion 6 Procedure for solving equations using a matrix inverse Let A be an n n matrix Let B be an arbitrary matrix Then we ask whether there are matrices C and D of suitable dimension such that

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 37 AC B D A B (46) In the first case the matrix B must have n rows, while in the second B must have n columns Then we have the following theorem Theorem 2 If A 0, then: AC B C A B D A B D B A (47) As an example, solve the following system of equations using theorem 2 We can write the system as 3 x + 5 x 2 4 8 x 2 x 2 22 (48) Ax b [ ] 3 5 A, x 8 2 x A b A [ 4 22 We can compute A from equation 0 We can now compute A b as follows A (3)( 2) (8)(5) [ ] 2 5 46 8 3 [ 2 ] 5 46 46 A b [ 2 46 8 46 3 46 5 46 8 46 3 46 [ 38 ] 46 46 46 [ ] x, b x 2 ] [ ] 4 22 [ ] 2 5 8 3 (49) (50) ] [ ] [ 4 28 46 + ] 0 46 2 22 46 66 46 [ ] (5) 3 62 Some example systems Solve each of the following using matrix inversion 3x + x 2 6 5x x 2 2 4x + x 2 0 3x + 2 x 2 0 (52) (53)

38 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 4x + 3 x 2 0 x + 2 x 2 5 (54) 4x + 3 x 2 x + 2 x 2 4 (55) 7 A general formula for the inverse of a matrix using adjoint matrices 7 The adjoint of a matrix The adjoint of the matrix A denoted adj (A) or A + is the transpose of the matrix obtained from A by replacing each element a ij by its cofactor A ij For example consider the matrix A 2 3 0 5 2 (56) 0 4 Now find the cofactor of each element For example the cofactor of the in the upper left hand corner is computed as A ( ) (+) 5 2 0 4 20 (57) Similarly the cofactor of a 23 2 is given by A 23 ( ) (2+3) 2 0 The entire matrix of cofactors is given by A A 2 A n A 2 A 22 A 2n A n A n2 A nn We can then compute the adjoint by taking the transpose A + A A 2 A n A 2 A 22 A 2n A n A n2 A nn 20 8 2 2 5 2 5 2 (58) 20 2 5 8 2 (59) 2 5 20 2 5 8 2 2 5 (60) 72 Finding an inverse matrix using adjoints For a square nonsingular matrix A, its inverse is given by A A A+ (6)

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 39 73 An example of finding the inverse using the adjoint First find the cofactor matrix A 2 3 0 5 2 0 4 (62) 20 2 5 A ij 8 2 2 5 Now find the transpose of A ij The determinant of A is Putting it all together we obtain A + 20 8 2 2 (63) 5 2 5 A 9 A 9 20 8 20/9 8/9 /9 2 2 2/9 /9 2/9 (64) 5 2 5 5/9 2/9 5/9 We can check the answer as follows AA 2 3 20/9 8/9 /9 0 5 2 2/9 /9 2/9 9/9 0 0 0 9/9 0 I (65) 0 4 5/9 2/9 5/9 0 0 9/9 Consider the general equation system 8 The general form of Cramer s rule where A is n n, x is nx, and b is nx A x b (66) Let D j denote the determinant formed from A by replacing the j th column with the column vector b Thus a a 2 a j b a j+ a n D j a 2 a 22 a 2 j b 2 a 2 j+ a 2n (67) a n a n2 a n j b n a n j+ a nn is Then the general system of n equations in n unknowns has unique solution if A 0 The solution x D A, x 2 D 2 A,, x n D n A (68)

40 MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 9 A note on homogeneous systems of equations The homogeneous system of n equations in n unknowns or Ax 0 (69) a x + a 2 x 2 + + a n x n b a 2 x + a 22 x 2 + + a 2n x n b 2 a 3 x + a 32 x 2 + + a 3n x n b 3 + + + (70) a m x + a m2 x 2 + + a mn x n b m has a nontrivial solution (a trivial solution has x 0, x 2 0, x n 0 if and only if A 0 Lecture Questions : Provide two examples of linear systems that have no solution and explain why 2: Provide two examples of linear systems that have exactly one solution and explain why 3: Provide two examples of linear systems that have exactly infinitely many solutions and explain why Provide at least one of these where the number of equations and unknowns is the same 4: Explain why if a linear system has exactly one solution, the coefficient matrix A must have at least as many rows as columns Give examples of why this is the case 5: If a system of linear equations has more unknowns than equations, it must either have no solution or infinitely many solutions Explain why Provide at least two examples