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# Class Note for CSCI 124 at GW

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Date Created: 02/07/15
Determinants and Matrix Inverses Notes for CSCi 124 PoorVi L Vora Consider the set of 71 linear equations in n unknowns 17 2 mm 111951 0112962 aina n bi 121951 0122962 1mm 52 M1951 an z ammn bm They may be written as a matrix equation AX b where an 112 ali aln 021 an 02139 027 A anl anZ am ann bi 52 b b VL and 951 902 X mn DetenninantsVomGWU 2 For example for the set of equations 3u5vw 4 6u11v4w 1 73uU72w 6 3 5 1 A 6 11 73 1 72 4 b 1 6 and 951 X 2 903 One way of solving the equations would be to multiply each side by the inverse of the matrix A A le A lb x A lb While gaussian elimination in general is more ef cient than nding A 1 because you also have to multiply A 1 with b it is sometimes useful to determine A 1 if one expects to solve several sets of equations with the same value of A as in the examples discussed in class computerized tomography and image deblurring An ef cient way of doing so is to use a technique similar to gaussian elimination known as the GaussJordan method 1 GaussJordan Method for Matrix Inverse The matrix to be inverted A is written along side the identity matrix of size n denoted In and gaussian elimination applied ALL Consider the example above 3 5 1 A 6 11 73 1 72 The matrix is 3 5 1 1 0 0 6 11 4 0 1 0 73 1 72 0 0 1 Detenni39nantsVomGWU 3 and is reduced to 3 5 1 1 0 0 0 1 2 E2 1 0 0 6 E1 1 0 1 and then to 3 5 1 1 0 0 0 1 2 E2 1 0 0 0 E13 13 E6 1 Thus the numbers below the diagonal of the matrix 3 5 1 0 1 2 0 0 E 13 are now zero Now we work on the numbers abave the diagonal working from the last row up 351100 0127210 0071313761 becomes 35 021731173 0100E 0071313761 and 71179 30 02E 0100 0071313 76 1 Now we diVide each row so as to get the identity in the rst three columns the other three provide the inverse 1 2 0 1 0 0 E E 6 1 0 0 1 71 E 713 Thus 2 711 79 3 39 39 71 1 2 A 0 E E 1 i 1 DetenninantsVomGWU 4 And it can be checked that AA 1 A lA 13 and also that A lb 117 4 the solution also obtained using gaussian elimination An explicit solution for a set of linear equations can be determined without resorting to gaussian elimination or the GaussJordan inverse though it is a far less ef cient method It is known as Cramer s Rule and uses the notion of the determinant of a matrix Before explaining Cramer s rule we explain the determinant 2 Determinants We start with examples The determinant of a matrix of size 2 x 2 is simple all a12 021 022 111022 7 112an where lAl denotes the determinant of A also often denoted DetA The determinant may be thought of as the volume of a parallelopiped solid whose sides are the column vectors of the ma trix In this case as the matrix is square matrix of size 2 the volume is the area of the parallelogram de ned by the two columns Example 1 Find the determinants of the matrices a 3 4 27 7 20 7 5 9 b 5 4 20 20 40 75 4 Now consider a matrix of size 3 x 3 an 112 113 122 123 121 123 121 122 121 122 123 an i 112 113 132 ass 0131 ass 0131 132 131 132 ass 022 a2 a32 a3 Here is termed the minar of an Notice that an is multiplied by a negative sign Example 2 Find the determinants of the matrices a HgtCHOJ 4 1 9 1 318727410741107364872472672 2 2 DetenninantsVomGWU 5 b 5 4 71 75 4 72 544747567171071240742258 3 2 1 We now de ne the determinant of a square matrix of size n x n determinants are not de ned for matrices that are not square an a12 aln 021 a22 a2 anl anZ ann allMinora11ialgMinor012a13Minora137a14Minora147l1na7mMinora7m TL 27li1a1iMinora1i i1 where M Modalj is the determinant of the matrix formed by deleting the row and column con taining aij all 012 a1j1 a1j1 am 021 022 1241 a2j1 a2 Minor lu 7 aiild 04712 aiiljil 0471441 aiilm l 7 7 ai11 ai12 ai1j71 ai1j1 ai1n am an2 amjil amjurl Mm Note that a matrix is invertible if and only if its determinant is nonzero 3 Cramer s Rule for the Solution of Linear Equations Recall our example for gaussian elimination 3u5vw 47777l 6ullv4w 177772 73uU72w 677776 DeterminantsVoraGWU 6 Instead of using gaussian elimination we use Cramer s rule to solve these Cramer s rule applied here is 4 5 1 1 11 4 U7 6 1 2 i 472274757272411766 77104130765773971 3 5 1 37227475712121633 77839 739 6 11 4 73 1 72 3 4 1 6 1 4 U7 73 6 2 i37272474712121363 777839773971 3 5 1 739 739 739 6 11 4 73 1 72 3 5 4 6 11 1 wi 3 1 6 i36671753634633719571951567156774 3 5 1 739 739 739 6 11 4 73 1 72 In general for the matrix equations AX b as de ned at the beginning of this set of notes Cramer s rule provides the following solutions when DetA 74 O and no solutions otherwise Thus it only provides a solution when a unique one exists A unique solution for a matrix equation of the form AX 6 exists if and only if DetA 74 O 7 D6tltAi 7 DetA mi where A is as de ned earlier and A1 is A with the 2 column replaced with the colunm of the rhs of the equation an a12 011121 b1 011141 am 121 122 12121 52 012141 aZn A1 anl anZ amiil bn ani1 ann 4 Exercises 1 Evaluate the determinants of the following matrices Detenni39nantsVomGWU 7 a 2 9 4 18 b 3 72 l7 51 c 3 2 5 4 3 1 2 73 76 d 1 1 71 2 1 1 71 71 2 2 Use Cramer s rule to determine the unique solution if one exists to the following sets of linear equations a m 1 3y 7 4 2m y i 16 b m y 7 z 7 4 2m y z 16 7x 7 y 22 i 0 You can use your answer to 1d here the following are the same equations you solved using gaussian elimination c 2U 61 7 4w 18 Bu 101 7 9w 27 5U 7 101 10w 710 d Bu 81 1 4w 7 Bu 101 7 2w 73 6U 7 101 10w 78 e u U 1 2w 2 5U 101 1 9w 4 2U 7 81 1 3w 13 3 Use the GaussJordan method to determine the inverse of the following matrices DetenninantsVomGWU a 1 2 71 A 2 1 0 71 1 72 b 1 2 72 A 1 1 0 71 1 72 c 1 0 71 A 1 1 0 1 1 71 You can also use GaussJordan to solve the equations in 2 and compare results Detenninan tsVoraGWU 9 5 Solutions 1 Evaluate the determinants of the following matrices a 2 9 418 2X1879X40 b 3 72 7 5 3x5772x729 c 3 2 5 4 3 1 2 73 76 3718 3 7 2726 5718 745 52 7 90 783 d 1 1 71 2 1 1 71 71 2 21 41 2 1 1 2 Use Cramer s rule to determine the unique solution if one exists to the following sets of linear equations a 4 3 161 744 44 3513 75E 14 216 8 y 13 7573 Detenni39nan tsVoraGWU 1 0 i 7 8 Solution is z 7 3 y 7 73 b m y 7 z 7 4 2m y z 16 7x 7 y 22 i 0 You can use your answer to 1d here 4 1 71 16 1 1 7 0 71 2 7421713270717167074 3 71 71 Denominator is the answer from 1d 1 4 71 2 16 1 1 0 2 13270744171016 y 4 71 71 1 1 4 2 1 16 1 1 0 1016710164721 z 4 71 71 Solutionism y 2 4 0 Solution is u 21 310 1 d u 7312 1andw 2 is the solution e u 11 71 andw 1 is the solution 3 Use the GaussJordan method to determine the inverse of the following matrices 1271 A 210 71172 1271100 A113 210010 71172001 1271100 073 27210 0 373101 11 DetenninantsVomGWU 0 1 7210 71 2 3 0 4 07171 0 0 0 0 72100 1 2 i2 1 1 i1 1 0 i2 1 2 1 1 0 0 1 0 i2 0 0 1 711 A113 0 2721 7110 71 1 0 2 0 1 7110 272 71 7231 1 0 2 2 0 i1 i2 1 1 i2 0 0 0 i1 3 0 2 12 DetenninantsVomGWU 1 1 0 0 1 1 12 2 32 71 i1 010 001 1 1 0 i1 1 0 0 1 1 0 0 1 0 1 1 i1 0 0 1 A113 17110 1 71 10011 01071 0 0 1 0 1 0

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