maths-matrix(2)

2011-05-09 1:26 am

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 (b=-1;x=t, y=2t+1, z=t-2)(ii)a=-1[b=5/3;x=t, y=(5-6t)/5, z=-(3t+10)/3](b part 我已做了) c) Is there a realsolution(x0,y0,z0) of equations(D) satisfying (x^2)+(y^2)+(z^2)=1?explain.......x-z=2(D) {3x-y-z=1.......x-y+z=-3

回答 (1)

myisland8132

2011-05-09 2:07 am

✔ 最佳答案

(c)

This is the case of (a) = 1 in (b) with solution

(x,y,z) = (2 + t, 5 + 2t, t)

if there a real solution(x0,y0,z0) of equations(D) satisfying (x^2)+(y^2)+(z^2)=1

(2 + t)^2 + (5 + 2t)^2 + t^2 = 1

6t^2 + 24t + 28 = 0

3t^2 + 12t + 14 = 0

Discriminant = (12)(12) - (12)(14) = -24 < 0

So, there is no real value of t satisfies the equation and thus no real solution of (x,y,z)

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