1. Fi equation. nd a parametric representation of the solution set of the linear

Question

1. Fi equation. nd a parametric representation of the solution set of the linear -4x + 2y - 6z 1 Solve the system using Gaussian-Jordan elimination: 3. Determine the values of k such that the system of linear equations has (a) exactly one solution (b) no solution (c) infinitely many solutions. kx + y = 0 4. Verify (AB)"-BTA" if 2 11 A=1-2 11 and B=[ -1] 0 4 5. Find the inverse of the following matrix without using formula. 2 -3 1 -2 6. Find an LU-factorization of the following matrix. Write out elementary matrices and inverse of elementary matrices in the process of finding an LU-factorization: 0 3 -7 -16 11 1

Explanation / Answer

Answer :

1. For parametric form of -4x + 2y - 6z =1, first I need to take an initial solution, a trivial guess can be (0,1/2,0). Now I need two vectors in the plane. I can do this by finding two other points in the plane, and subtracting them from this one (the difference of two vectors points from one to the other, so if both points are in the plane their difference will point along it). I'll take the points (-1/4,0,0) and (0,0,-1/6). Notice the simple construction of all my points: set two variables to zero and find out what the third one should be. It's probably the easiest way to go.

So my vectors are going to be these two points minus the original one found.

1st vector => (-1/4,-1/2,0) and 2nd vector => (0,-1/2,-1/6)

Now any vector in the plane, when scaled, is still in the plane. So I can define my plane like this

(0,1/2,0) + (?1/4,?1/2,0)t + (0,-1/2,-1/6)s . Now simply splitting this up in terms of components (x,y,z) instead of points we get, x= -1/4 t , y = 1/2 -1/2 t - 1/2 s , z = -1/6 s.

2. Performing RRE operations on the 3x4 coefficient matrix.

we finally get

Therefore, solution set: x = 4 + 3z , y = 5 + 2z and z = any

3. Make coefficient matrix as in previous problem

On inspection we get that for k=1 and -1 the system has no solution since det of the coefficient matrix made up of only first two rows and colum from the previous matrix becomes zero and for no value the system has infinite solutions since for infinitely many solutions k/1 = 1/k = 0/1 = 0. Which cannot be true for any real k. For all other k values the system has unique solution.

4. Matrix Multiplication

5. Matrix inverse is given as A-1 = 1/det A * Cofactor matrix.

Here Cofactor matrix is

So inverse is and det A = -4 - (-3) = -1

x y z b 1 -2 1 -6 2 -3 0 -7 -1 3 -3 11

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.