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Problems #1,9 from pg. 47 are graded.

#1, pg.47: Determine if the system has a nontrivial solution: $$\left\{ \begin{array}{ll} 2x_1-5x_2+8x_3=0 \\ -2x_1 - 7x_2 + x_3 = 0 \\ 4x_1 + 2x_2 + 7x_3 = 0 \end{array} \right.$$ Solution: We see that this a homogeneous equation and so its solution is either trivial or nontrivial. We also know that in the case of a nontrivial solution, we musth ave a free variable in the solution. So we will set up an augmented matrix for this system and row reduce until the answer is apparent. Form the augmented matrix for this system compute $$\begin{array}{ll} \left[\begin{array}{llll} 2 & -5 & 8 & 0 \\ -2 & -7 & 1 & 0 \\ 4 & 2 & 7 & 0 \end{array}\right] &\stackrel{r_2^*=r_2+r_1}{\stackrel{r_3^*=r_4-2r_1}{\sim}} \left[\begin{array}{llll} 2 & -5 & 8 & 0 \\ 0 & -12 & 9 & 0 \\ 0 & 12 & -9 & 0 \end{array} \right] \\ &\stackrel{r_3^*=r_3+r_2}{\sim} \left[ \begin{array}{llll} 2 & -5 & 8 & 0 \\ 0 & -12 & 9 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{array}$$ Because of the third row of zeros, we will have a free variable in the solution vector $\vec{x}=\left[ \begin{array}{ll} x_1 \\ x_2 \\ x_3 \end{array}\right]$ and hence there will be a nontrivial solution to the system of equations.

#9, pg.47: Describe all solutions of $A\vec{x}=\vec{0}$ in parametric vector form, where $A = \left[ \begin{array}{llll} 3 & -6 & 6 \\ -2 & 4 & -2 \end{array} \right]$.

Solution: In this case we are forced to take $\vec{x} \in \mathbb{R}^{3 \times 1}$ so let $\vec{x}=\left[ \begin{array}{llll} x_1 \\ x_2 \\ x_3 \end{array} \right]$. The equation $A\vec{x}=\vec{0}$ is equivalent to the solution of the system whose augmented matrix is $$\begin{array}{ll} \left[ \begin{array}{llll} 3 & -6 & 6 & 0 \\ -2 & 4 & -2 & 0 \end{array} \right] &\stackrel{r_2^*=r_2+\frac{2}{3}r_1}{\sim} \left[ \begin{array}{llll} 3 & -6 & 6 & 0 \\ 0 & 0 & 2 & 0 \end{array} \right] \\ &\stackrel{r_1^* = r_1-3r_2}{\sim} \left[ \begin{array}{llll} 3 & -6 & 0 & 0 \\ 0 & 0 & 2 & 0. \end{array} \right] \\ &\stackrel{r_1^*=\frac{1}{3}r_1}{\stackrel{r_2^*=\frac{1}{2}r_2}{\sim}} \left[ \begin{array}{llll} 1 & -2 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \end{array}$$ This yields solution vector $$\vec{x} = \left[ \begin{array}{llll} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{llll} 2x_2 \\ x_2 \\ 0 \end{array} \right] = x_2 \left[ \begin{array}{llll} 2 \\ 1 \\ 0 \end{array} \right].$$ We see that our solution has a free variable and hence there exist nontrivial solutions to the homogeneous equation, in particular, one solution for every posible value of $x_2$.