Answer to Question #346040 in Linear Algebra for Beza

Question #346040

Find the solution set of x1+2x2-3x3+x4=0

3X1-X2+5X3-X4=0

2X1+X2+X4=0


1
Expert's answer
2022-05-30T16:28:36-0400

Write down the augmented matrix


A=(123103151021010)A=\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 3 & -1 & 5 & -1 &| & 0 \\ 2 & 1 & 0 & 1 &| & 0 \\ \end{pmatrix}

R2=R23R1R_2=R_2-3R_1


(1231007144021010)\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & -7 & 14 & -4 &| & 0 \\ 2 & 1 & 0 & 1 &| & 0 \\ \end{pmatrix}

R3=R32R1R_3=R_3-2R_1


(1231007144003610)\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & -7 & 14 & -4 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}

R2=R2/7R_2=-R_2/7


(123100124/7003610)\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}

R1=R12R2R_1=R_1-2R_2


(1011/700124/7003610)\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}

R3=R3+3R2R_3=R_3+3R_2


(1011/700124/700005/70)\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 5/7 &| & 0 \\ \end{pmatrix}

R3=7R3/5R_3=7R_3/5


(1011/700124/7000010)\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}


R1=R1+R3/7R_1=R_1+R_3/7


(101000124/7000010)\begin{pmatrix} 1 & 0 & 1 & 0 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}

R2=R24R3/7R_2=R_2-4R_3/7


(101000120000010)\begin{pmatrix} 1 & 0 & 1 & 0 &| & 0 \\ 0 & 1 & -2 & 0 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}

Take x3=t,tR.x_3=t, t\in \R.

Then x1=t,x2=2t,x3=t,x4=0,tRx_1=-t, x_2=2t, x_3=t, x_4=0, t\in \R



{t,2t,t,0},tR\{-t, 2t, t, 0\}, t\in \R


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