lecture4-09[1]

lecture4-09[1] - Math 18.02 (Spring 2009): Lecture 4 Square...

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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 4 Square systems. Equations of planes February 10 Reading Material: From Simmons : 18.4. From Course Notes M. Last time: Matrices. Inverse Matrices Today: Square systems. Word problem. How many solutions? Equations of planes 2 Recall: Matrix Inversion Given a n n matrix A we recall the sequence of steps you need to obtain its inverse, assuming of course that | A | 6 = 0: B B @ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 1 C C A B B @ m 11 m 12 m 13 m 21 m 22 m 23 m 31 m 32 m 33 1 C C A B B @ m 11- m 12 m 13- m 21 m 22- m 23 m 31- m 32 m 33 1 C C A B B @ m 11- m 21 m 31- m 12 m 22- m 32 m 13- m 32 m 33 1 C C A A M C C T minors cofactors transpose Then A- 1 = 1 | A | C T . Remark. Why does it work? . 1 | A | ( a 11 m 11- a 12 m 12 + a 13 m 13 ) AA- 1 = A ( 1 | A | C T ) = 1- 1 | A | ( a 21 m 11- a 22 m 12 + a 23 m 13 ) cofator exp. of | A | where first row of A is replaced with 2 nd row 2 identical rows! Then det = 0 . More precisely lets consider for example the entry at place 11 in AA- 1 . Since we require that AA- 1 = I n , as indicated in the picture, for this entry we must have 1 | A | [ a 11 m 11- a 12 m 12 + a 13 m 13 ] = 1 . 1 This is indeed correct since a 11 m 11- a 12 m 12 + a 13 m 13 is exactly | A | computed using the expansion with respect to the first row. On the other hand for the element at position 21 in AA- 1 we must have 1 | A | [ a 21 m 11- a 22 m 12 + a 23 m 13 ] = 0 . Now notice that a 21 m 11- a 22 m 12 + a 23 m 13 is the determinant, also obtained by expanding with respect to the first rwo, but of the matrix obtained from A by replacing the first row by the second one. This determinant is in fact zero since the matrix has two equal rows....
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This note was uploaded on 03/01/2009 for the course 18 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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lecture4-09[1] - Math 18.02 (Spring 2009): Lecture 4 Square...

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