Question #169663

solve a system step-by-step of linear equations using Gaussian elimination.

2x + 7y + z = 1

x + 3y - z = 2

x + 7y + 12z = 45



1
Expert's answer
2021-03-09T03:55:07-0500
{2x+7y+z=1x+3yz=2x+7y+12z=45\begin{cases} 2x + 7y + z = 1 \\ x + 3y - z = 2 \\ x + 7y + 12z = 45 \\ \end{cases}

(27111312171245)R3+(R2)(27111312041343)R2+(12R1)(27110123232041343)\begin{pmatrix} 2 & 7 & 1 &| 1 \\ 1 & 3 &-1 & |2 \\ 1 & 7 & 12 &|45 \\ \end{pmatrix} \underrightarrow{R_3 + (-R_2)} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 1 & 3 &-1 & |2 \\ 0 & 4 & 13 &|43 \\ \end{pmatrix} \underrightarrow{R_2 + (-\frac{1}{2}R_1)} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 0 & -\frac{1}{2} &-\frac{3}{2} & |\frac{3}{2} \\ 0 & 4 & 13 &|43 \\ \end{pmatrix} \\


R3+8R2(2711012323200155)12R1;2R2;(1721212013300155)\\ \\ \underrightarrow{R_3 + 8R_2} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 0 & -\frac{1}{2} &-\frac{3}{2} & |\frac{3}{2} \\ 0 & 0 & 1 &|55 \\ \end{pmatrix} \underrightarrow{\frac{1}{2}R_1 ; -2R_2;} \begin{pmatrix} 1 & \frac{7}{2} & \frac{1}{2} &| \frac{1}{2} \\ 0 & 1 & 3 & | -3 \\ 0 & 0 & 1 &|55 \\ \end{pmatrix} \Longrightarrow


{x+72y+12z=12y+3z=3z=55\Longrightarrow \begin{cases} x + \frac{7}{2}y + \frac{1}{2}z = \frac{1}{2} \\ y + 3z = -3 \\ z = 55 \\ \end{cases}z=55;y+355=3;y=168;x+72(168)+1255=12;x588+27,5=0,5;x=561;z = 55;\\ y + 3 \cdot 55 = -3; \\ y = -168; \\ x + \frac {7}{2} \cdot (-168) + \frac{1}{2} \cdot 55 = \frac{1}{2}; \\ x -588 + 27,5 = 0,5; \\ x= 561;

Answer: x=561;y=168;z=55.x = 561; y = -168; z = 55.


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