Answer on Question #85777 – Math – Linear Algebra
Question
Solve the system of equations
3 x + 2 y + 4 z = 7 3x + 2y + 4z = 7 3 x + 2 y + 4 z = 7 2 x + y + z = 1 2x + y + z = 1 2 x + y + z = 1 x + 3 y + 5 z = 2 x + 3y + 5z = 2 x + 3 y + 5 z = 2
with partial pivoting. Store the multipliers and also write the pivoting vectors.
Solution
3 x + 2 y + 4 z = 7 2 x + y + z = 1 x + 3 y + 5 z = 2 \begin{array}{l}
3x + 2y + 4z = 7 \\
2x + y + z = 1 \\
x + 3y + 5z = 2 \\
\end{array} 3 x + 2 y + 4 z = 7 2 x + y + z = 1 x + 3 y + 5 z = 2 ( 3 ∗ 2 4 7 2 1 1 1 1 3 5 2 ) → 1 2 R 1 ( 1 ∗ 2 / 3 4 / 3 7 / 3 2 1 1 1 1 3 5 2 ) \left( \begin{array}{ccccc}
3^* & 2 & 4 & 7 \\
2 & 1 & 1 & 1 \\
1 & 3 & 5 & 2
\end{array} \right) \xrightarrow{ \frac{1}{2} R_1 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
2 & 1 & 1 & 1 \\
1 & 3 & 5 & 2
\end{array} \right) ⎝ ⎛ 3 ∗ 2 1 2 1 3 4 1 5 7 1 2 ⎠ ⎞ 2 1 R 1 ⎝ ⎛ 1 ∗ 2 1 2/3 1 3 4/3 1 5 7/3 1 2 ⎠ ⎞ ( 1 ∗ 2 / 3 4 / 3 7 / 3 2 1 1 1 1 3 5 2 ) → R 2 − 2 R 1 ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 − 1 / 3 − 5 / 3 − 11 / 3 1 3 5 2 ) \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
2 & 1 & 1 & 1 \\
1 & 3 & 5 & 2
\end{array} \right) \xrightarrow{R_2 - 2R_1 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & -1/3 & -5/3 & -11/3 \\
1 & 3 & 5 & 2
\end{array} \right) ⎝ ⎛ 1 ∗ 2 1 2/3 1 3 4/3 1 5 7/3 1 2 ⎠ ⎞ R 2 − 2 R 1 ⎝ ⎛ 1 ∗ 0 1 2/3 − 1/3 3 4/3 − 5/3 5 7/3 − 11/3 2 ⎠ ⎞ ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 − 1 / 3 − 5 / 3 − 11 / 3 1 3 5 2 ) → R 3 − R 1 ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 − 1 / 3 − 5 / 3 − 11 / 3 0 7 / 3 11 / 3 − 1 / 3 ) \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & -1/3 & -5/3 & -11/3 \\
1 & 3 & 5 & 2
\end{array} \right) \xrightarrow{R_3 - R_1 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & -1/3 & -5/3 & -11/3 \\
0 & 7/3 & 11/3 & -1/3
\end{array} \right) ⎝ ⎛ 1 ∗ 0 1 2/3 − 1/3 3 4/3 − 5/3 5 7/3 − 11/3 2 ⎠ ⎞ R 3 − R 1 ⎝ ⎛ 1 ∗ 0 0 2/3 − 1/3 7/3 4/3 − 5/3 11/3 7/3 − 11/3 − 1/3 ⎠ ⎞ ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 − 1 / 3 − 5 / 3 − 11 / 3 0 7 / 3 11 / 3 − 1 / 3 ) → − 3 R 2 ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 1 5 11 0 7 / 3 11 / 3 − 1 / 3 ) \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & -1/3 & -5/3 & -11/3 \\
0 & 7/3 & 11/3 & -1/3
\end{array} \right) \xrightarrow{-3R_2 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & 1 & 5 & 11 \\
0 & 7/3 & 11/3 & -1/3
\end{array} \right) ⎝ ⎛ 1 ∗ 0 0 2/3 − 1/3 7/3 4/3 − 5/3 11/3 7/3 − 11/3 − 1/3 ⎠ ⎞ − 3 R 2 ⎝ ⎛ 1 ∗ 0 0 2/3 1 7/3 4/3 5 11/3 7/3 11 − 1/3 ⎠ ⎞ ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 1 5 11 0 7 / 3 11 / 3 − 1 / 3 ) → R 3 − 7 3 R 2 ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 1 5 11 0 0 − 8 − 26 ) \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & 1 & 5 & 11 \\
0 & 7/3 & 11/3 & -1/3
\end{array} \right) \xrightarrow{R_3 - \frac{7}{3} R_2 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & 1 & 5 & 11 \\
0 & 0 & -8 & -26
\end{array} \right) ⎝ ⎛ 1 ∗ 0 0 2/3 1 7/3 4/3 5 11/3 7/3 11 − 1/3 ⎠ ⎞ R 3 − 3 7 R 2 ⎝ ⎛ 1 ∗ 0 0 2/3 1 0 4/3 5 − 8 7/3 11 − 26 ⎠ ⎞ ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 1 5 11 0 0 − 8 − 26 ) → − 1 8 R 3 ( 1 ∗ 2 / 3 4 / 3 7 / 3 0 1 5 11 0 0 1 13 / 4 ) \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & 1 & 5 & 11 \\
0 & 0 & -8 & -26
\end{array} \right) \xrightarrow{-\frac{1}{8} R_3 } \left( \begin{array}{ccccc}
1^* & 2/3 & 4/3 & 7/3 \\
0 & 1 & 5 & 11 \\
0 & 0 & 1 & 13/4
\end{array} \right) ⎝ ⎛ 1 ∗ 0 0 2/3 1 0 4/3 5 − 8 7/3 11 − 26 ⎠ ⎞ − 8 1 R 3 ⎝ ⎛ 1 ∗ 0 0 2/3 1 0 4/3 5 1 7/3 11 13/4 ⎠ ⎞ z = 13 4 y = 11 − 5 z = 11 − 5 ( 13 4 ) = − 21 4 x = 7 3 − 2 3 y − 4 3 z = 7 3 − 2 3 ( − 21 4 ) − 4 3 ( 13 4 ) = 3 2 \begin{array}{l}
z = \frac{13}{4} \\
y = 11 - 5z = 11 - 5\left(\frac{13}{4}\right) = -\frac{21}{4} \\
x = \frac{7}{3} - \frac{2}{3}y - \frac{4}{3}z = \frac{7}{3} - \frac{2}{3}\left(-\frac{21}{4}\right) - \frac{4}{3}\left(\frac{13}{4}\right) = \frac{3}{2}
\end{array} z = 4 13 y = 11 − 5 z = 11 − 5 ( 4 13 ) = − 4 21 x = 3 7 − 3 2 y − 3 4 z = 3 7 − 3 2 ( − 4 21 ) − 3 4 ( 4 13 ) = 2 3
Answer:
( x , y , z ) = ( 3 2 , − 21 4 , 13 4 ) . (x, y, z) = \left(\frac {3}{2}, - \frac {21}{4}, \frac {13}{4}\right). ( x , y , z ) = ( 2 3 , − 4 21 , 4 13 ) .
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#340153 on Dec 2023