Chapter 8: Chapter 8

Show that the following system has more than one solution. $$ 3 x-y+7 z=0 $$ $$ \begin{aligned} &2 \mathrm{x}-\mathrm{y}+4 \mathrm{z}=(1 / 2) \\ &\mathrm{x}-\mathrm{y}+\mathrm{z}=1 \\ &6 \mathrm{x}-4 \mathrm{y}+10 \mathrm{z}=3 \end{aligned} $$

Find the solutions of the following systems and describe the solutions in geometric terms. (a) \(-2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}=0\) \(2 x-y-3 z=0\) \(-6 x+3 y+9 z=0\) (b) \(2 x+5 y+z=0\) \(\mathrm{x}-2 \mathrm{y}+\mathrm{z} \quad=0\) \(3 x+3 y+2 z=0\) (c) \(x-y+z \quad=0\) \(2 x-y+z \quad=0\) \(\mathrm{x}+\mathrm{y}+\mathrm{z} \quad=0\)

Find the necessary and sufficient conditions for the existence of a solution to the following system. $$ \begin{aligned} x+y+2 z &=a_{1} \\ -2 x-z &=a_{2} \\ x+3 y+5 z &=a_{3} \end{aligned} $$

Determine the values of a so that the following system of (a) no solution, (b) more than one solution, (c) a unique solution. $$ \begin{array}{r} x+y-z=1 \\ 2 x+3 y+a z=3 \\ x+a y+3 z=2 \end{array} $$

Solve the following system $\mathrm{x}_{1}-2 \mathrm{x}_{2}-3 \mathrm{x}_{3}=3$ \(2 \mathrm{x}_{1}-\mathrm{x}_{2}-4 \mathrm{x}_{3}=7\) \(3 \mathrm{x}_{1}-3 \mathrm{x}_{2}-5 \mathrm{x}_{3}=8\)

Find the solution set of the following system of equations: \(2 \mathrm{x}_{1}+\mathrm{x}_{2}-4 \mathrm{x}_{3}=8\) \(3 \mathrm{x}_{1}-\mathrm{x}_{2}+2 \mathrm{x}_{3}=-1\)

Consider the following nonhomogeneous system of linear equations. $2 \mathrm{x}+\mathrm{y}-3 \mathrm{z}=1$ \(3 x+2 y-2 z=2\) \(\mathrm{x}+\mathrm{y}+\mathrm{z} \quad=1\) Show that (i) any two solutions to the system (1) differ by a vector which is a solution to the homogeneous system \(2 \mathrm{x}+\mathrm{y}-3 \mathrm{z}=0\) \(3 x+2 y-2 z=0\) \(\mathrm{x}+\mathrm{y}+\mathrm{z} \quad=0\) (ii) the sum of a solution to (1) and a solution to (2) gives a solution to (1).

Solve the following systems of equations. $\begin{array}{ll}\text { (1) } & 4 \mathrm{x}_{1}-3 \mathrm{x}_{2}+\mathrm{x}_{3}=-1 \\ & \mathrm{x}_{1}+5 \mathrm{x}_{2}-2 \mathrm{x}_{3}=2 \\ & \mathrm{x}_{1}+2 \mathrm{x}_{2}=0 \\\ \text { (2) } & 2 \mathrm{x}_{1}+3 \mathrm{x}_{2}+\mathrm{x}_{3}-4 \mathrm{x}_{4}=0 \\ & \mathrm{x}_{1}-5 \mathrm{x}_{2}-3 \mathrm{x}_{3}+2 \mathrm{x}_{4}=0 \\ & 5 \mathrm{x}_{1}+2 \mathrm{x}_{2}-\mathrm{x}_{4}=0 & \\\ & 2 \mathrm{x}_{1}-9 \mathrm{x}_{2}-5 \mathrm{x}_{3}+9 \mathrm{x}_{4}=0 \\\ \text { (3) } & 8 \mathrm{x}_{1}-2 \mathrm{x}_{2}+4 \mathrm{x}_{3}+3 \mathrm{x}_{4}+\mathrm{x}_{5} & =2 \\ & \mathrm{x}_{2}-4 \mathrm{x}_{3}+\mathrm{x}_{4} & -2 \mathrm{x}_{5} & =-100 \\ & 2 \mathrm{x}_{1}+\mathrm{x}_{2}-\quad 4 \mathrm{x}_{4} & =1\end{array}$

Show that the system of linear equations over the rational number field \(Q\) $2 \mathrm{x}_{1}+6 \mathrm{x}_{2}-\mathrm{x}_{3}+\mathrm{x}_{4}=2$ \(\mathrm{x}_{1}+3 \mathrm{x}_{2}+\mathrm{x}_{3} \quad=5\) \(-x_{1}-3 x_{2}-x_{3}\) \(=0\) has no solution.

If the method of Gauss elimination corresponds in its final form to an echelon matrix, what is the matrix analogue of the Gauss-Jordan method for solving linear systems of equations? Explain by example.