Answer to Question #194546 in Linear Algebra for Zeeshan

Question #194546

Using gaussian elimination method find all solutions to the following system of linear equations.

2x₂ + 3x₃ + 4x₄ = 1

x₁ - 3x₂ + 4x₃ + 5x₄ = 2

-3x₁ + 10x₂ - 6x₃ -7x₄ = -4


1
Expert's answer
2021-05-19T12:58:05-0400
[0234   11345   231067   4]\begin{bmatrix} 0 & 2 & 3 & 4 & \ \ \ 1 \\ 1 & -3 &4 & 5 & \ \ \ 2 \\ -3 & 10 & -6 & -7 & \ \ \ -4 \\ \end{bmatrix}



Swap rows 1 and 2


[1345   20234   131067   4]\begin{bmatrix} 1 & -3 & 4 & 5 & \ \ \ 2 \\ 0 & 2 & 3 & 4 & \ \ \ 1 \\ -3 & 10 & -6 & -7 & \ \ \ -4 \\ \end{bmatrix}

R3=R3+3R1R_3=R_3+3R_1


[1345   20234   10168   2]\begin{bmatrix} 1 & -3 & 4 & 5 & \ \ \ 2 \\ 0 & 2 & 3 & 4 & \ \ \ 1 \\ 0 & 1 & 6 & 8 & \ \ \ 2 \\ \end{bmatrix}



R2=R2/2R_2=R_2/2


[1345   2013/22   1/20168   2]\begin{bmatrix} 1 & -3 & 4 & 5 & \ \ \ 2 \\ 0 & 1 & 3/2 & 2 & \ \ \ 1/2 \\ 0 & 1 & 6 & 8 & \ \ \ 2 \\ \end{bmatrix}


R1=R1+3R2R_1=R_1+3R_2


[1017/211   7/2013/22   1/20168   2]\begin{bmatrix} 1 & 0 & 17/2 & 11 & \ \ \ 7/ 2 \\ 0 & 1 & 3/2 & 2 & \ \ \ 1/2 \\ 0 & 1 & 6 & 8 & \ \ \ 2 \\ \end{bmatrix}



R3=R3R2R_3=R_3-R_2

[1017/211   7/2013/22   1/2009/26   3/2]\begin{bmatrix} 1 & 0 & 17/2 & 11 & \ \ \ 7/ 2 \\ 0 & 1 & 3/2 & 2 & \ \ \ 1/2 \\ 0 & 0 & 9/2 & 6 & \ \ \ 3/2 \\ \end{bmatrix}


R3=2R3/9R_3=2R_3/9


[1017/211   7/2013/22   1/20014/3   1/3]\begin{bmatrix} 1 & 0 & 17/2 & 11 & \ \ \ 7/ 2 \\ 0 & 1 & 3/2 & 2 & \ \ \ 1/2 \\ 0 & 0 & 1 & 4/3 & \ \ \ 1/3 \\ \end{bmatrix}


R1=R117R3/2R_1=R_1-17R_3/2


[1001/3   2/3013/22   1/20014/3   1/3]\begin{bmatrix} 1 & 0 & 0 & -1/3 & \ \ \ 2/ 3 \\ 0 & 1 & 3/2 & 2 & \ \ \ 1/2 \\ 0 & 0 & 1 & 4/3 & \ \ \ 1/3 \\ \end{bmatrix}


R2=R23R3/2R_2=R_2-3R_3/2

[1001/3   2/30100   00014/3   1/3]\begin{bmatrix} 1 & 0 & 0 & -1/3 & \ \ \ 2/ 3 \\ 0 & 1 & 0 & 0 & \ \ \ 0 \\ 0 & 0 & 1 & 4/3 & \ \ \ 1/3 \\ \end{bmatrix}




x113x4=23x2=0x3+43x4=13\begin{matrix} x_1-\dfrac{1}{3}x_4=\dfrac{2}{3} \\ \\ x_2=0 \\ \\ x_3+\dfrac{4}{3}x_4=\dfrac{1}{3} \\ \\ \end{matrix}

Solution set:

x1=23+13x4x2=0x3=1343x4x4, free\begin{matrix} x_1=\dfrac{2}{3} +\dfrac{1}{3}x_4 \\ \\ x_2=0 \\ \\ x_3=\dfrac{1}{3}-\dfrac{4}{3}x_4 \\ \\ x_4,\ free \\ \\ \end{matrix}

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