Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Elimination leads to an upper triangular system then

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e. Then there will be infinitely many solutions (two parallel lines become one single line). The equation y D 1 comes from elimination (subtract x C y D 5 from x C 2y D 6). Then x D 4 and 5x 4y D c D 16. 1 (a) Another solution is 2 .x C X; y C Y; z C Z/. (b) If 25 planes meet at two points, they meet along the whole line through those two points. Elimination leads to an upper triangular system; then comes back substitution. 2x C 3y C z D 8 xD2 y C 3z D 4 gives y D 1 If a zero is at the start of row 2 or 3, 8z D 8 z D 1 that avoids a row operation. 2x 3y D 3 2x 3y D3 2x 3y D 3 xD3 y C z D 1 and 4x 5y C z D 7 gives y C z D 1 and y D 1 2x y 3z D 5 5z D 0 zD0 2y C 3z D 2 Subtract 2  row 1 from row 2, subtract 1  row 1 from row 3, subtract 2  row 2 from row 3 Subtract 2 times row 1 from row 2 to reach .d 10/y z D 2. Equation (3) is y z D 3. If d D 10 exchange rows 2 and 3. If d D 11 the system becomes singular. The second pivot position will contain 2 b . If b D 2 we exchange with row 3. If b D 1 (singular case) the second equation is y z D 0. A solution is .1; 1; 1/. 0x C 0y C 2z D 4 Exchange 0x C 3y C 4z D 4 Example of x C 2y C 2z D 5 but then x C 2y C 2z D 5 (a) 2 exchanges (b) 0x C 3y C 4z D 6 break down 0x C 3y C 4z D 6 (exchange 1 and 2, then 2 and 3) (rows 1 and 3 are not consistent) If row 1 D row 2, then row 2 is zero after the first step; exchange the zero row with row 3 and there is no third pivot. If column 2 D column 1, then column 2 has no pivot. Example x C 2y C 3z D 0, 4x C 8y C 12z D 0, 5x C 10y C 15z D 0 has 9 different coefficients but rows 2 and 3 become 0 D 0: infinitely many solutions. Row 2 becomes 3y 4z D 5, then row 3 becomes .q C 4/z D t 5. If q D 4 the system is singular—no third pivot. Then if t D 5 the third equation is 0 D 0. Choosing z D 1 the equation 3y 4z D 5 gives y D 3 and equation 1 gives x D 9. Singular if row 3 is a combination of rows 1 and 2. From the end view, the three planes form a triangle. This happens if rows 1 C 2 D row 3 on the left side but not the right side: x C y C z D 0, x 2y z D 1, 2x y D 4. No parallel planes but still no solution. 345 3 4 5 21 (a) Pivots 2; 2 ; 3 ; 4 in the equations 2x C y D 0; 2 y C z D 0; 3 z C t D 0; 4 t D 5 after elimination. Back substitution gives t D 4; z D 3; y D 2; x D 1. (b) If the off-diagonal entries change from C1 to 1, the pivots are the same. The solution is .1; 2; 3; 4/ instead of . 1; 2; 3; 4/. 6 22 The fifth pivot is 5 for both matrices (1’s or nC1 . n 1’s off the diagonal). The nth pivot is Solutions to Exercises 11 23 If ordinary elimination leads to x C y D 1 and 2y D 3, the original second equation 24 25 26 27 28 29 30 31 32 could be 2y C `.x C y/ D 3 C ` for any `. Then ` will be the multiplier to reach 2y D 3.   a2 Elimination fails on if a D 2 or a D 0. aa a D 2 (equal columns), a D 4 (equal rows), a D 0 (zero column). Solvable for s D 10 (add the two pairs of equations to get a C b C c C d on the left sides, 12 and 2 C s on the right sides). The four equations for a; b; c; d are singular! Two 2 3 2 3 1100 1 100     13 04 1 1 07 61 0 1 07 60 solutio...
View Full Document

This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

Ask a homework question - tutors are online