Answer on Question#54581 - <math> - <linear algebra="">
Solve the linear system
{ 2 x + y − 3 z = 5 3 x − 2 y − 2 z = 3 5 x − 3 y − 2 z = 16 \left\{ \begin{array}{l} 2x + y - 3z = 5 \\ 3x - 2y - 2z = 3 \\ 5x - 3y - 2z = 16 \end{array} \right. ⎩ ⎨ ⎧ 2 x + y − 3 z = 5 3 x − 2 y − 2 z = 3 5 x − 3 y − 2 z = 16
Solution. Let's construct an augmented matrix with y y y x x x z z z
[ 1 2 − 3 − 2 3 − 2 − 3 5 − 2 ] 5 3 16 \left[ \begin{array}{ccc} 1 & 2 & -3 \\ -2 & 3 & -2 \\ -3 & 5 & -2 \end{array} \right] \begin{array}{c} 5 \\ 3 \\ 16 \end{array} ⎣ ⎡ 1 − 2 − 3 2 3 5 − 3 − 2 − 2 ⎦ ⎤ 5 3 16
Let's solve it by Gaussian elimination.
[ 1 2 − 3 5 − 2 3 − 2 3 − 3 5 − 2 16 ] ∼ [ 1 2 − 3 5 0 7 − 8 13 0 11 − 11 31 ] ∼ [ 1 2 − 3 5 0 1 − 8 / 7 13 / 7 0 11 − 11 31 ] ∼ [ 1 2 − 3 5 0 1 − 8 / 7 13 / 7 0 0 − 77 + 88 7 217 − 143 7 ] ∼ ∼ [ 1 2 − 3 5 0 1 − 8 / 7 13 / 7 0 0 11 / 7 74 / 7 ] ∼ [ 1 2 − 3 5 0 1 − 8 / 7 13 / 7 0 0 1 74 / 11 ] ∼ [ 1 2 − 3 5 0 1 0 143 + 592 77 0 0 1 74 / 11 ] ∼ [ 1 2 − 3 5 0 1 0 1 0 0 1 74 / 11 ] ∼ ∼ [ 1 0 − 3 55 − 210 11 0 1 0 105 / 11 0 0 1 74 / 11 ] ∼ [ 1 0 − 3 − 155 11 0 1 0 105 / 11 0 0 1 74 / 11 ] ∼ [ 1 0 0 − 155 + 222 11 0 1 0 105 / 11 0 0 1 74 / 11 ] ∼ [ 1 0 0 67 / 11 0 1 0 105 / 11 0 0 1 74 / 11 ] \begin{array}{l}
\left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ -2 & 3 & -2 & 3 \\ -3 & 5 & -2 & 16 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 7 & -8 & 13 \\ 0 & 11 & -11 & 31 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & -8/7 & 13/7 \\ 0 & 11 & -11 & 31 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & -8/7 & 13/7 \\ 0 & 0 & \frac{-77 + 88}{7} & \frac{217 - 143}{7} \end{array} \right] \sim \\
\sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & -8/7 & 13/7 \\ 0 & 0 & 11/7 & 74/7 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & -8/7 & 13/7 \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & 0 & \frac{143 + 592}{77} \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 2 & -3 & 5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \\
\sim \left[ \begin{array}{cccc} 1 & 0 & -3 & \frac{55 - 210}{11} \\ 0 & 1 & 0 & 105/11 \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 0 & -3 & \frac{-155}{11} \\ 0 & 1 & 0 & 105/11 \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 0 & 0 & \frac{-155 + 222}{11} \\ 0 & 1 & 0 & 105/11 \\ 0 & 0 & 1 & 74/11 \end{array} \right] \sim \left[ \begin{array}{cccc} 1 & 0 & 0 & 67/11 \\ 0 & 1 & 0 & 105/11 \\ 0 & 0 & 1 & 74/11 \end{array} \right]
\end{array} ⎣ ⎡ 1 − 2 − 3 2 3 5 − 3 − 2 − 2 5 3 16 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 7 11 − 3 − 8 − 11 5 13 31 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 1 11 − 3 − 8/7 − 11 5 13/7 31 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 1 0 − 3 − 8/7 7 − 77 + 88 5 13/7 7 217 − 143 ⎦ ⎤ ∼ ∼ ⎣ ⎡ 1 0 0 2 1 0 − 3 − 8/7 11/7 5 13/7 74/7 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 1 0 − 3 − 8/7 1 5 13/7 74/11 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 1 0 − 3 0 1 5 77 143 + 592 74/11 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 2 1 0 − 3 0 1 5 1 74/11 ⎦ ⎤ ∼ ∼ ⎣ ⎡ 1 0 0 0 1 0 − 3 0 1 11 55 − 210 105/11 74/11 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 0 1 0 − 3 0 1 11 − 155 105/11 74/11 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 0 1 0 0 0 1 11 − 155 + 222 105/11 74/11 ⎦ ⎤ ∼ ⎣ ⎡ 1 0 0 0 1 0 0 0 1 67/11 105/11 74/11 ⎦ ⎤
Thus, we obtain the solution:
{ x = 105 11 y = 67 11 z = 74 11 \left\{ \begin{array}{l} x = \frac{105}{11} \\ y = \frac{67}{11} \\ z = \frac{74}{11} \end{array} \right. ⎩ ⎨ ⎧ x = 11 105 y = 11 67 z = 11 74
Answer: ( 105 11 , 67 11 , 74 11 ) \left(\frac{105}{11}, \frac{67}{11}, \frac{74}{11}\right) ( 11 105 , 11 67 , 11 74 )
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#340153 on Dec 2023