solve a system step-by-step of linear equations using Gaussian elimination.
2x + 7y + z = 1
x + 3y - z = 2
x + 7y + 12z = 45
(271∣113−1∣21712∣45)R3+(−R2)→(271∣113−1∣20413∣43)R2+(−12R1)→(271∣10−12−32∣320413∣43)\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} \\⎝⎛2117371−112∣1∣2∣45⎠⎞R3+(−R2)⎝⎛2107341−113∣1∣2∣43⎠⎞R2+(−21R1)⎝⎛2007−2141−2313∣1∣23∣43⎠⎞
R3+8R2→(271∣10−12−32∣32001∣55)12R1;−2R2;→(17212∣12013∣−3001∣55)⟹\\ \\ \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} \LongrightarrowR3+8R2⎝⎛2007−2101−231∣1∣23∣55⎠⎞21R1;−2R2;⎝⎛10027102131∣21∣−3∣55⎠⎞⟹
Answer: x=561;y=−168;z=55.x = 561; y = -168; z = 55.x=561;y=−168;z=55.
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