Linear Algebra Practical Class, Week 21

Question One

Question

Find the solution set which corresponds to the following row echelon forms:

1. $M_1=\left[\begin{array}{ccccc}1& -2& -1& |& 0\\ 0& 2& 1& |& 0\end{array}\right]$,

2. $M_2=\left[\begin{array}{cccccc}1& 0& 1& 2& |& 0\\ 0& 3& -1& -1& |& 0\end{array}\right]$.

Solution

$$\begin{array}{rl}{R}_2\mapsto R_2+R_1& ⟺\left[\begin{array}{ccccc}1& 0& 0& |& 0\\ 0& 1& \frac{1}{2}& |& 0\end{array}\right]\\ & \\ & ⟺\{\begin{array}{l}x=0\\ y+\frac{1}{2}z=0\end{array}\\ & ⟺z=-2y\\ & ⟺S=\{\left[\begin{array}{c}0\\ y\\ -2y\end{array}\right]\in {\mathbb{R}}^3:y\in \mathbb{R}\}\\ & ⟺S=\text{span}(\left[\begin{array}{c}0\\ 1\\ -2\end{array}\right])\end{array}$$

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$$\begin{array}{r}...\end{array}$$

Question Two

Question

Solve the system with three equations and four unknowns using Gauss-Jordan elimination, and write its set of solutions:

$$\{\begin{array}{l}{x}_1+2x_2-2x_3+2x_4=1\\ 2x_1-4x_2+4x_3+6x_4=0\\ x_1+2x_2-2x_3+3x_4=0\end{array}$$

Solution

$$\begin{array}{rl}\{\begin{array}{l}{x}_1+2x_2-2x_3+2x_4=1\\ 2x_1-4x_2+4x_3+6x_4=0\\ x_1+2x_2-2x_3+3x_4=0\end{array}& ⟺\left[\begin{array}{cccccc}1& 2& -2& 2& |& 1\\ 2& -4& 4& 6& |& 0\\ 1& 2& -2& 3& |& 0\end{array}\right]\\ & ⟺R_1\mapsto R_1-R_3\\ & ⟺\left[\begin{array}{cccccc}0& 0& 0& -1& |& 1\\ 2& -4& 4& 6& |& 0\\ 1& 2& -2& 3& |& 0\end{array}\right]\\ & ⟺R_2\mapsto \frac{R_2+2R_3}{4}\\ & ⟺\left[\begin{array}{cccccc}0& 0& 0& -1& |& 1\\ 1& 0& 0& 3& |& 0\\ 1& 2& -2& 3& |& 0\end{array}\right]\\ & ⟺\{\begin{array}{l}-x_4=1\\ x_1+3x_4=0\\ x_1+2x_2-3x_3+3x_4=0\end{array}\\ & ⟺x_1=3,x_4=-1,x_2=\frac{3}{2}{x}_3\\ & ⟺S=\{\left[\begin{array}{c}3\\ \frac{3}{2}{x}_3\\ x_3\\ -1\end{array}\right]\in {\mathbb{R}}^4:x_3\in \mathbb{R}\}\end{array}$$

Question Three

Question

Let ${\mathbf{v}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\mathbf{v}}_2=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],{\mathbf{v}}_3=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ be vectors in ${\mathbb{R}}^3$.

Using Gauss-Jordan elimination:

1. Show that the vectors are linearly dependent.

2. Determine whether the vector $\mathbf{u}=\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]$ belongs to the span of the vectors ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$.

Solution

$$\begin{array}{rl}\left[\begin{array}{ccccc}1& 1& 1& |& 0\\ 0& 1& 1& |& 0\\ 0& 0& 1& |& 0\end{array}\right]& ⟺R_1\mapsto R_1-R_2,R_2\mapsto R_2-R_3\\ & ⟺\left[\begin{array}{ccccc}1& 0& 0& |& 0\\ 0& 1& 0& |& 0\\ 0& 0& 1& |& 0\end{array}\right]\\ & ⟺\{\begin{array}{l}x=0\\ y=0\\ z=0\end{array}\\ & \therefore \text{linearly dependent}\end{array}$$

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$$\begin{array}{rl}\left[\begin{array}{ccccc}1& 1& 1& |& 0\\ 0& 1& 1& |& 1\\ 0& 0& 1& |& 1\end{array}\right]& ⟺R_2\mapsto R_2-R_3\\ & ⟺\left[\begin{array}{ccccc}1& 1& 1& |& 0\\ 0& 1& 0& |& 0\\ 0& 0& 1& |& 1\end{array}\right]\\ & ⟺\{\begin{array}{l}x+y+z=0\\ y=0\\ z=0\end{array}\\ & ⟺x=y=z=0\end{array}$$

Question Four

Question

Considering the following matrices, denote which are in row-echelon form and which of those are in reduced row-echelon form:

1. $B_1=\left[\begin{array}{ccc}-1& 0& 1\\ 0& 2& 1\\ 0& 0& 5\\ 0& 0& -1\end{array}\right]$,

2. $B_2=\left[\begin{array}{cccc}1& 2& 0& 5\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$,

3. $B_3=\left[\begin{array}{cc}1& -1\\ 0& 1\\ 0& 0\end{array}\right]$,

4. $B_4=\left[\begin{array}{ccccc}1& 0& 0& -2& 5\\ 0& 1& 0& 9& -2\\ 0& 0& 1& 4& 3\end{array}\right]$.

Solution

Row Echelon Form: $B_2,B_3,B_4$. Reduced Row Echelon Form: $B_2,B_4$.