Chapter 1.4, Problem 38E

To determine

Does the system of linear equation have more than one solution? Justify your answer..

Answer to Problem 38E

Yes the system of linear equation has more than one solution.

Explanation of Solution

Given information: Three equations are given:

  $\begin{array}{l}4x+y+z=0(i)\\ 2x+\frac{1}{2}y-3z=0(ii)\\ -x-\frac{1}{4}y-z=0(iii)\end{array}$

Calculation:

  $\begin{array}{l}4x+y+z=0(i)\\ 2x+\frac{1}{2}y-3z=0(ii)\\ -x-\frac{1}{4}y-z=0(iii)\end{array}$

Equation $(ii)$ multiplies by $2$ , equation $(iii)$ multiplies by $4$ and add we get:

  $\begin{array}{l}4x+y-6z=0\\ -4x-y-4z=0\\ \overline{-10z=0}\\ z=0\end{array}$

Put $z=0$ in equation $(i)\mathrm{and}(ii)$ we get:

  $\begin{array}{l}4x+y+0=0\\ 4x+y=0(iv)\\ \mathrm{In}\mathrm{equation}(ii)\\ 2x+\frac{1}{2}y-3(0)=0\\ \frac{4x+y}{2}=0\\ 4x+y=0(v)\end{array}$

Equation $(iv)\mathrm{and}(v)$ are same therefore system has many solutions.

  $\begin{array}{l}y=-4x\\ \mathrm{let}x=a\\ \mathrm{Then}y=-4b\end{array}$

Therefore this system has many solutions of type $(x,y,z)=(a,-4a,0)$ .