Answer to Question #169663 in Linear Algebra for william

Question

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

2x + 7y + z = 1

x + 3y - z = 2

x + 7y + 12z = 45

Answer 1

$$\begin{cases} 2x + 7y + z = 1 \\ x + 3y - z = 2 \\ x + 7y + 12z = 45 \\ \end{cases}$$

$\begin{pmatrix} 2 & 7 & 1 &| 1 \\ 1 & 3 &-1 & |2 \\ 1 & 7 & 12 &|45 \\ \end{pmatrix} \underrightarrow{R_3 + (-R_2)} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 1 & 3 &-1 & |2 \\ 0 & 4 & 13 &|43 \\ \end{pmatrix} \underrightarrow{R_2 + (-\frac{1}{2}R_1)} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 0 & -\frac{1}{2} &-\frac{3}{2} & |\frac{3}{2} \\ 0 & 4 & 13 &|43 \\ \end{pmatrix} \\$

$\\ \\ \underrightarrow{R_3 + 8R_2} \begin{pmatrix} 2 & 7 & 1 &| 1 \\ 0 & -\frac{1}{2} &-\frac{3}{2} & |\frac{3}{2} \\ 0 & 0 & 1 &|55 \\ \end{pmatrix} \underrightarrow{\frac{1}{2}R_1 ; -2R_2;} \begin{pmatrix} 1 & \frac{7}{2} & \frac{1}{2} &| \frac{1}{2} \\ 0 & 1 & 3 & | -3 \\ 0 & 0 & 1 &|55 \\ \end{pmatrix} \Longrightarrow$

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

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