Question #120102
Solve the following system of linear equations
x+ y- 3z+ 2w = 0
2x -y +2z- 3w =0
3x- 2y+ z -4w =0
-4x +y -3z +w =0
1
Expert's answer
2020-06-04T19:35:48-0400
A=(1132212332144131),X=(xyzw),B=(0000)A=\begin{pmatrix} 1 & 1 & -3 & 2 \\ 2 & -1 & 2 & -3 \\ 3 & -2 & 1 & -4 \\ -4 & 1 & -3 & 1 \end{pmatrix}, X=\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}, B=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}

AX=BAX=B

R2=R2(2)R1R_2=R_2-(2)R_1


(1132038732144131)\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 3 & -2 & 1 & -4 \\ -4 & 1 & -3 & 1 \end{pmatrix}

R3=R3(3)R1R_3=R_3-(3)R_1


(113203870510104131)\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 0 & -5 & 10 & -10 \\ -4 & 1 & -3 & 1 \end{pmatrix}

R4=R4+(4)R1R_4=R_4+(4)R_1


(1132038705101005159)\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}

R2=R2/(3)R_2=R_2/(-3)


(1132018/37/305101005159)\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 &1 & -8/3 & 7/3 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}

R1=R1R2R_1=R_1-R_2


(101/31/3018/37/305101005159)\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}

R3=R3+(5)R2R_3=R_3+(5)R_2


(101/31/3018/37/30010/35/305159)\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & -10/3 & 5/3 \\ 0 & 5 & -15 & 9 \end{pmatrix}

R4=R4(5)R2R_4=R_4-(5)R_2


(101/31/3018/37/30010/35/3005/38/3)\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & -10/3 & 5/3 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}

R3=(3/10)R3R_3=(-3/10)R_3


(101/31/3018/37/30011/2005/38/3)\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}

R1=R1+(1/3)R3R_1=R_1+(1/3)R_3


(1001/2018/37/30011/2005/38/3)\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}

R2=R2+(8/3)R3R_2=R_2+(8/3)R_3


(1001/201010011/2005/38/3)\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}

R4=R4+(5/3)R3R_4=R_4+(5/3)R_3


(1001/201010011/20007/2)\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & 0 & -7/2 \end{pmatrix}

R4=(2/7)R4R_4=(-2/7)R_4


(1001/201010011/20001)\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & 0 & 1 \end{pmatrix}

R1=R1+(1/2)R4R_1=R_1+(1/2)R_4


(100001010011/20001)\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & 0 & 1 \end{pmatrix}

R2=R2R4R_2=R_2-R_4


(100001000011/20001)\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & 0 & 1 \end{pmatrix}

R3+R3+(1/2)R4R_3+R_3+(1/2)R_4


(1000010000100001)\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

(1000010000100001)(xyzw)=(0000)\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}

x=0x=0

y=0y=0

z=0z=0

w=0w=0


(0,0,0,0)(0, 0,0,0)



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