Question #298576

solve the linear system by gaussian elimination

 x − y + 2z − w = −1,

2x + y − 2z − 2w = −2 ,

−x + 2y − 4z + w = 1 ,

3x − 3w = −3


1
Expert's answer
2022-02-18T01:43:53-0500

Augmented matrix


[11211212221241130033]\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 2 & 1 & -2 & -2 & & -2\\ -1 & 2 & -4 & 1 & & 1 \\ 3 & 0 & 0 & -3 & & -3 \\ \end{bmatrix}

R2=R22R1R_2=R_2-2R_1


[11211036001241130033]\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ -1 & 2 & -4 & 1 & & 1 \\ 3 & 0 & 0 & -3 & & -3 \\ \end{bmatrix}

R3=R3+R1R_3=R_3+R_1


[11211036000120030033]\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 3 & 0 & 0 & -3 & & -3 \\ \end{bmatrix}

R4=R43R1R_4=R_4-3R_1


[11211036000120003600]\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}

R2=R2/3R_2=R_2/3


[11211012000120003600]\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}

R1=R1+R2R_1=R_1+R_2


[10011012000120003600]\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}

R3=R3R2R_3=R_3-R_2


[10011012000000003600]\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 0 & 0 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}

R4=R43R2R_4=R_4-3R_2


[10011012000000000000]\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 0 & 0 & 0 & & 0 \\ 0 & 0 &0 & 0 & & 0 \\ \end{bmatrix}

If w=t,tR,z=s,sR,w=t, t\in \R, z=s, s\in \R, then x=1+t,y=2s,z=s,w=t,t,sR.x=-1+t, y=2s, z=s, w=t, t,s\in \R.


The linear system has infinitely many solutions


(1+t,2s,s,t), t,sR(-1+t, 2s, s, t), \ t,s\in \R

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