Question #202118

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)].


1
Expert's answer
2021-06-03T10:04:20-0400

1. Let


c1u1+c2u2+c3u3=0c_1\vec u_1+c_2\vec u_2+c_3\vec u_3=0


Then


3c1+2c2+c3=03c_1+2c_2+c_3=0

4c1c2+6c3=04c_1-c_2+6c_3=0

3c28c3=03c_2-8c_3=0

c1+5c22c3=0c_1+5c_2-2c_3=0

(3210416003801520)\begin{pmatrix} 3 & 2 & 1 & 0\\ 4 & -1 & 6 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}

R1=R13R_1=\dfrac{R_1}{3}


(12/31/30416003801520)\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 4 & -1 & 6 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}

R2=R24R1R_2=R_2-4R_1


(12/31/30011/314/3003801520)\begin{pmatrix} 1 & 2/3 & 1/3 & 0\\ 0 & -11/3 & 14/3 & 0 \\ 0 & 3 & -8& 0 \\ 1 & 5 & -2 & 0 \\ \end{pmatrix}

R4=R4R1R_4=R_4-R_1


(12/31/30011/314/300380013/37/30)\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}

R2=311R2R_2=-\dfrac{3}{11}R_2


(12/31/300114/1100380013/37/30)\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}

R1=R12R23R_1=R_1-\dfrac{2R_2}{3}


(1013/1100114/1100380013/37/30)\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 3 & -8& 0 \\ 0 & 13/3 & -7/3 & 0 \\ \end{pmatrix}

R3=R33R2R_3=R_3-3R_2


(1013/1100114/1100046/110013/37/30)\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}

R4=R413R23R_4=R_4-\dfrac{13R_2}{3}


(1013/1100114/1100046/1100035/110)\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & -46/11 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}

R3=11R346R_3=-\dfrac{11R_3}{46}


(1013/1100114/11000100035/110)\begin{pmatrix} 1 & 0 & 13/11 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}

R1=R113R311R_1=R_1-\dfrac{13R_3}{11}


(10000114/11000100035/110)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -14/11 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}

R2=R2+14R311R_2=R_2+\dfrac{14R_3}{11}


(1000010000100035/110)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 35/11 & 0 \\ \end{pmatrix}

R4=R435R311R_4=R_4-\dfrac{35R_3}{11}


(1000010000100000)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}

c1=c2=c3=0c_1=c_2=c_3=0

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


2. Let


c1v1+c2v2+c3v3+c4v4=0c_1\vec v_1+c_2\vec v_2+c_3\vec v_3+c_4\vec v_4=0


Then


2c1+7c4=02c_1+7c_4=0

2c1+8c4=02c_1+8c_4=0

2c1+9c4=02c_1+9c_4=0

2c1+c3=02c_1+c_3=0(20070200802009020100)\begin{pmatrix} 2 & 0 & 0 & 7 & 0\\ 2 & 0 & 0 & 8 & 0\\ 2 & 0 & 0 & 9 & 0 \\ 2 & 0 & 1 & 0 & 0 \\ \end{pmatrix}

R1=R12R_1=\dfrac{R_1}{2}


(1007/20200802009020100)\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}

R2=R22R1R_2=R_2-2R_1


(1007/20000102009020100)\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}

R3=R32R1R_3=R_3-2R_1


(1007/20000100002020100)\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}

R4=R42R1R_4=R_4-2R_1


(1007/20000100002000170)\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


(1007/20001700002000010)\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}

R3=R32R_3=\dfrac{R_3}{2}

(1007/20001700001000010)\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}


R1=R17R32R_1=R_1-\dfrac{7R_3}{2}

(10000001700001000010)\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & -7 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}

R2=R2+7R3R_2=R_2+7R_3

(10000001000001000010)\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{pmatrix}

R4=R4R3R_4=R_4-R_3

(10000001000001000000)\begin{pmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}

c1=c3=c4=0,c2Rc_1=c_3=c_4=0, c_2\in \R

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


3. Let


c1w1+c2w2+c3w3=0c_1\vec w_1+c_2\vec w_2+c_3\vec w_3=0


Then


6c1+2c24c3=06c_1+2c_2-4c_3=0

2c24c3=02c_2-4c_3=0

c1+5c24c3=0-c_1+5c_2-4c_3=0

3c14c3=03c_1-4c_3=0(6240024015403040)\begin{pmatrix} 6 & 2 & -4 & 0\\ 0 & 2 & -4 & 0 \\ -1 & 5 & -4& 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}

R1=R16R_1=\dfrac{R_1}{6}

(11/32/30024015403040)\begin{pmatrix} 1 & 1/3 & -2/3 & 0\\ 0 & 2 & -4 & 0 \\ -1 & 5 & -4& 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}

R3=R3+R1R_3=R_3+R_1

(11/32/300240016/314/303040)\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 2 & -4 & 0 \\ 0 & 16/3 & -14/3 & 0 \\ 3 & 0 & -4 & 0 \\ \end{pmatrix}

R4=R43R1R_4=R_4-3R_1

(11/32/300240016/314/300120)\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 2 & -4 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}

R2=R22R_2=\dfrac{R_2}{2}

(11/32/300120016/314/300120)\begin{pmatrix} 1 & 1/3 & -2/3& 0\\ 0 & 1 & -2 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}

R1=R1R23R_1=R_1-\dfrac{R_2}{3}

(10000120016/314/300120)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 &16/3 & -14/3 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}

R3=R316R23R_3=R_3-\dfrac{16R_2}{3}

(1000012000600120)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & -1 & -2 & 0 \\ \end{pmatrix}

R4=R4+R2R_4=R_4+R_2

(1000012000600040)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}


R3=R36R_3=\dfrac{R_3}{6}

(1000012000100040)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}

R2=R2+2R3R_2=R_2+2R_3

(1000010000100040)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -4 & 0 \\ \end{pmatrix}

R4=R4+4R3R_4=R_4+4R_3

(1000010000100000)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}

c1=c2=c3=0c_1=c_2=c_3=0

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



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