Answer to Question #202118 in Linear Algebra for rama

Question

Show that the given set of vectors are L.I or L.D. (i) u = [(3; 4; 0; 1); (2; -1; 3; 5); (1; 6; -8; -2)].

(ii) v = [(2; 0; 0; 7); (2; 0; 0; 8); (2; 0; 0; 9); (2; 0; 1; 0)] (iii) w = [(6; 0; -1; 3); (2; 2; 5; 0); (-4; -4; -4; -4)].

Answer 1

1. Let

$$c_1\vec u_1+c_2\vec u_2+c_3\vec u_3=0$$

Then

$$3c_1+2c_2+c_3=0$$

$$4c_1-c_2+6c_3=0$$

$$3c_2-8c_3=0$$

$$c_1+5c_2-2c_3=0$$

$$\begin{pmatrix} 3 & 2 & 1 & 0\\ 4 & -1 & 6 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}$$

$R_1=\dfrac{R_1}{3}$

$$\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 4 & -1 & 6 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}$$

$R_2=R_2-4R_1$

$$\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 0 & -11/3 & 14/3 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}$$

$R_4=R_4-R_1$

$$\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 0 & -11/3 & 14/3 & 0 \\ 0 & 3 & -8& 0 \\ 0 & 13/3 & -7/3 & 0 \\ \end{pmatrix}$$

$R_2=-\dfrac{3}{11}R_2$

$$\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 3 & -8& 0 \\ 0 & 13/3 & -7/3 & 0 \\ \end{pmatrix}$$

$R_1=R_1-\dfrac{2R_2}{3}$

$$\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 3 & -8& 0 \\ 0 & 13/3 & -7/3 & 0 \\ \end{pmatrix}$$

$R_3=R_3-3R_2$

$$\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & -46/11 & 0 \\ 0 & 13/3 & -7/3 & 0 \\ \end{pmatrix}$$

$R_4=R_4-\dfrac{13R_2}{3}$

$$\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & -46/11 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}$$

$R_3=-\dfrac{11R_3}{46}$

$$\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}$$

$R_1=R_1-\dfrac{13R_3}{11}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}$$

$R_2=R_2+\dfrac{14R_3}{11}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}$$

$R_4=R_4-\dfrac{35R_3}{11}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

$c_1=c_2=c_3=0$

The vectors are linear independent (L.I.).

2. Let

$$c_1\vec v_1+c_2\vec v_2+c_3\vec v_3+c_4\vec v_4=0$$

Then

$$2c_1+7c_4=0$$

$$2c_1+8c_4=0$$

$$2c_1+9c_4=0$$

$$2c_1+c_3=0$$ $$\begin{pmatrix} 2 & 0 & 0 & 7 & 0\\ 2 & 0 & 0 & 8 & 0\\ 2 & 0 & 0 & 9 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ \end{pmatrix}$$

$R_1=\dfrac{R_1}{2}$

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 2 & 0 & 0 & 8 & 0\\ 2 & 0 & 0 & 9 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ \end{pmatrix}$$

$R_2=R_2-2R_1$

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 0 & 0 & 0 & 1 & 0\\ 2 & 0 & 0 & 9 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ \end{pmatrix}$$

$R_3=R_3-2R_1$

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 2 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ \end{pmatrix}$$

$R_4=R_4-2R_1$

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 1 & -7 & 0 \\ \end{pmatrix}$$

Swap rows 2 and 4

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 0 & 0 & 1 & -7 & 0\\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$$

$R_3=\dfrac{R_3}{2}$

$$\begin{pmatrix} 1 & 0 & 0 & 7/2 & 0\\ 0 & 0 & 1 & -7 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$$

$R_1=R_1-\dfrac{7R_3}{2}$

$$\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & -7 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$$

$R_2=R_2+7R_3$

$$\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}$$

$R_4=R_4-R_3$

$$\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

$c_1=c_3=c_4=0, c_2\in \R$

The vectors are linear dependent (L.D.).

3. Let

$$c_1\vec w_1+c_2\vec w_2+c_3\vec w_3=0$$

Then

$$6c_1+2c_2-4c_3=0$$

$$2c_2-4c_3=0$$

$$-c_1+5c_2-4c_3=0$$

$$3c_1-4c_3=0$$ $$\begin{pmatrix} 6 & 2 & -4 & 0\\ 0 & 2 & -4 & 0 \\ -1 & 5 & -4& 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_1=\dfrac{R_1}{6}$

$$\begin{pmatrix} 1 & 1/3 & -2/3 & 0\\ 0 & 2 & -4 & 0 \\ -1 & 5 & -4& 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_3=R_3+R_1$

$$\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 2 & -4 & 0 \\ 0 & 16/3 & -14/3 & 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_4=R_4-3R_1$

$$\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 2 & -4 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}$$

$R_2=\dfrac{R_2}{2}$

$$\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 1 & -2 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}$$

$R_1=R_1-\dfrac{R_2}{3}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}$$

$R_3=R_3-\dfrac{16R_2}{3}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}$$

$R_4=R_4+R_2$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_3=\dfrac{R_3}{6}$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_2=R_2+2R_3$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}$$

$R_4=R_4+4R_3$

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

$c_1=c_2=c_3=0$

The vectors are linear independent (L.I.).