Introduction to Linear algebra-Strang-Solutions-Manual_ver13

6 4 their components still add to 1 8 3 6 6 8

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Unformatted text preview: ions normally lie on a line. Solutions to Exercises 9 28 The row picture shows four lines in the 2D plane. The column picture is in four- dimensional space. No solution unless the right side is a combination of the two columns.   :7 :65 29 u2 D and u3 D . The components add to 1. They are always positive. :3 :35 u7 ; v7 ; w7 are all close to .:6; :4/. Their components still add to 1.        :8 :3 :6 :6 :8 :3 30 D D steady state s. No change when multiplied by . :2 :7 :4 :4 :2 :7 " #" # 834 5Cu 5 uCv 5 v 5 5 C u C v ; M3 .1; 1; 1/ D .15; 15; 15/; 31 M D 1 5 9 D 5 u v 672 5Cv 5Cu v 5 u M4 .1; 1; 1; 1/ D .34; 34; 34; 34/ because 1 C 2 C    C 16 D 136 which is 4.34/. 32 A is singular when its third column w is a combination c u C d v of the first columns. A typical column picture has b outside the plane of u, v, w. A typical row picture has the intersection line of two planes parallel to the third plane. Then no solution. 33 w D .5; 7/ is 5u C 7v. Then Aw equals 5 times Au plus 7 times Av. 2 2 61 34 4 0 0 1 2 1 0 0 1 2 1 32 3 2 3 2 3 23 0 x1 1 x1 4 0 7 6 x2 7 6 2 7 6 x2 7 6 7 7 D has the solution 4 5 D 4 5. 1 5 4 x3 5 4 3 5 x3 8 2 x4 4 x4 6 35 x D .1; : : : ; 1/ gives S x D sum of each row D 1 C  C 9 D 45 for Sudoku matrices. 6 row orders .1; 2; 3/, .1; 3; 2/, .2; 1; 3/, .2; 3; 1/, .3; 1; 2/, .3; 2; 1/ are in Section 2.7. The same 6 permutations of blocks of rows produce Sudoku matrices, so 64 D 1296 orders of the 9 rows all stay Sudoku. (And also 1296 permutations of the 9 columns.) Problem Set 2.2, page 51 1 Multiply by `21 D 10 2 to circle are 2 and 6. 2 D 5 and subtract to find 2x C 3y D 14 and 6y D 6. The pivots 6y D 6 gives y D 1. Then 2x C 3y D 1 gives x D 2. Multiplying the right side .1; 11/ by 4 will multiply the solution by 4 to give the new solution .x; y/ D .8; 4/. (or add 1 ) times equation 1. The new second equation is 3y D 3. Then 2 y D 1 and x D 5. If the right side changes sign, so does the solution: .x; y/ D . 5; 1/. 3 Subtract 1 2 or .ad c a times equation 1. The new second pivot multiplying y is d b c/=a. Then y D .ag cf /=.ad b c/. 4 Subtract ` D .cb=a/ 5 6x C 4y is 2 times 3x C 2y . There is no solution unless the right side is 2  10 D 20. Then all the points on the line 3x C 2y D 10 are solutions, including .0; 5/ and .4; 1/. (The two lines in the row picture are the same line, containing all solutions). 6 Singular system if b D 4, because 4x C 8y is 2 times 2x C 4y . Then g D 32 makes the lines become the same: infinitely many solutions like .8; 0/ and .0; 4/. 7 If a D 2 elimination must fail (two parallel lines in the row picture). The equations have no solution. With a D 0, elimination will stop for a row exchange. Then 3y D gives y D 1 and 4x C 6y D 6 gives x D 3. 3 Solutions to Exercises 10 8 If k D 3 elimination must fail: no solution. If k D 9 10 11 12 13 14 15 16 17 18 19 20 3, elimination gives 0 D 0 in equation 2: infinitely many solutions. If k D 0 a row exchange is needed: one solution. On the left side, 6x 4y is 2 times .3x 2y/. Therefore we need b2 D 2b1 on the right sid...
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