5 5 5 5 10 a Multiply the first equation by –310 and subtract the result from the second equation to

79 14.3 14.35 14.4 14.45 14.5 14.55 14.6 14.65 14.7

9.75 10

10.25 b The determinant can be computed as 02 . 02 . 1 1 2 5 . 2 02 . 1 1 5 . = − − − = − − which is close to zero. c Because the lines have very similar slopes and the determinant is so small, you would expect that the system would be ill-conditioned d Multiply the first equation by 1.020.5 and subtract the result from the second equation to eliminate the x 1 term from the second equation, 58 . 04 . 5 . 9 5 . 2 2 1 = − = − x x x The second equation can be solved for 5 . 14 04 . 58 . 2 = = x This result can be substituted into the first equation which can be solved for 10 5 . 5 . 14 5 . 9 1 = + − = x e Multiply the first equation by 1.020.52 and subtract the result from the second equation to eliminate the x 1 term from the second equation, 16538 . 03846 . 5 . 9 52 . 2 2 1 − = − − = − x x x 80 The second equation can be solved for 3 . 4 03846 . 16538 . 2 = − − = x This result can be substituted into the first equation which can be solved for 10 52 . 3 . 4 5 . 9 1 − = + − = x Interpretation: The fact that a slight change in one of the coefficients results in a radically different solution illustrates that this system is very ill-conditioned.

8.7 a Multiply the first equation by –310 and subtract the result from the second equation to

eliminate the x 1 term from the second equation. Then, multiply the first equation by 110 and subtract the result from the third equation to eliminate the x 1 term from the third equation. 2 . 24 1 . 5 8 . 4 . 53 7 . 1 4 . 5 27 2 10 3 2 3 2 3 2 1 − = + − = + − = − + x x x x x x x Multiply the second equation by 0.8–5.4 and subtract the result from the third equation to eliminate the x 2 term from the third equation, 11111 . 32 5.351852 4 . 53 7 . 1 4 . 5 27 2 10 3 3 2 3 2 1 − = − = + − = − + x x x x x x Back substitution can then be used to determine the unknowns 5 . 10 8 2 6 27 8 4 . 5 6 7 . 1 4 . 53 6 351852 . 5 11111 . 32 1 2 3 = − − = = − − − − = − = − = x x x b Check: 81 5 . 21 6 5 8 5 . 5 . 61 6 2 8 6 5 . 3 27 6 8 2 5 . 10 − = − + + − = − + − − = − − +

8.8 a Pivoting is necessary, so switch the first and third rows,