Answer to Question #110818 in Linear Algebra for Hetisani Sewela

Question

Consider the following homogeneous system of linear equations
x + 4y + z = 0
4x + 13y + 7z = 0
7x + 22y + 3z = 0



.
(a) Determine the solution(s) of the above system by reducing the system to row echelon
form. (7)
(b) Is (2, 4, 2) a solution of the system? Give a reason for your answer. (3)
(c) State how you can use the determinant of the coefficient matrix of the above system to determine
if the system has nontrivial solutions or not. DO NOT EVALUATE THE ACTUAL DETER-
MINANT.

Answer 1

a) $\begin{cases} x + 4y + z = 0 \\ 4x + 13y + 7z = 0\\ 7x + 22y + 3z = 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 4y + z = 0 \\ 0x - 3y +3z = 0\\ 0x- 6y - 4z = 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 4y + z = 0 \\ 0x - 3y +3z = 0\\ 0x- 0y - 10z= 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 4y + z = 0 \\ 0x - 3y +3z = 0\\ z= 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 4y + z = 0 \\ 0x - 3y +0 = 0\\ z = 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 4y + z = 0 \\ y = 0\\ z = 0\\ \end{cases}$ $\implies$ $\begin{cases} x + 0 + 0 = 0 \\ y = 0\\ z = 0\\ \end{cases}$ $\implies$ $\begin{cases} x = 0 \\ y = 0\\ z = 0\\ \end{cases}$

b) Let's check whether (2, 4, 2) is a solution. If x = 2, y = 4, z = 2 we obtain the following:

from the first row: 2 + 4*4 + 2 $\neq$ 0. Thus, it is not the solution of the system (since it is not a solution for the first equation).

c) First of all let's consider the matrix of this system:

$\begin{pmatrix} 1 &4 & 1 \\ 4 & 13 & 7\\ 7 & 22 & 3 \end{pmatrix}$By the Cramer's rule:

If the determinant of this matrix is not equal to 0, we can state that system of equations has unique solution,and because of the system is homogeneous it is trivial.

Solution is not trivial if and only if the determinant is equal to 0.