maths-matrix急!!!!!!

2011-05-08 11:23 pm

Consider the system of linear equations(E),wherea and b are real numbers.2ax-2z=b+53x+by-az=-bx+by+az=3ba) Prove that a^2≠1 and b≠0 if (e) has a unique solution, and solve (e) accordingly.(a part 我已做了) b)For each of the following cases, find the value of b such that (e) hassolution. Solve (E) for the value of b obtained.(i)a=1(ii)a=-1(我未學gaussian elimination, gauss-jordan elimination,最好用Cramer's rule)

回答 (1)

myisland8132

2011-05-08 11:46 pm

✔ 最佳答案

(b)(i) a = 1

2x-2z=b+5...(1)3x+by-z=-b..(2)x+by+z=3b...(3)

(1) + (3): 3x + by - z = 4b + 5

So, 4b + 5 = -b => b = -1

Then x - z = 2, 3x - y - z = 1, x - y + z = -3

y = 5 + 2z, x = 2 + z

So, the solution is (x,y,z) = (2 + t, 5 + 2t, t)

(ii) a = -1

-2x-2z=b+5...(1)3x+by+z=-b..(2)x+by-z=3b...(3)

(3) - (2): -2x - 2z = 4b

So, 4b = b + 5 => b =5/3

Then -6x - 6z = 20, 9x + 5y + 3z = -5, 3x + 5y - 3z = 15

Let z = t, x = -(10 + 3t)/3,

y = (15 + 3z - 3x)/5 = (15 + 3t + 10 + 3t)/5

= (25 + 6t)/5

So, (x,y,z) = [-(10 + 3t)/3, (25 + 6t)/5, t]

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