Question

Solve the system of equations 3x+2y+4z=7
2x+y+z=1
x+3y+5z=2
with partial pivoting.Store the multipliers and also write the pivoting vectors.

Answer 1

Answer on Question #85777 – Math – Linear Algebra

Question

Solve the system of equations

$$3x + 2y + 4z = 7$$ $$2x + y + z = 1$$ $$x + 3y + 5z = 2$$

with partial pivoting. Store the multipliers and also write the pivoting vectors.

Solution

$$\begin{array}{l} 3x + 2y + 4z = 7 \\ 2x + y + z = 1 \\ x + 3y + 5z = 2 \\ \end{array}$$ $$\left( \begin{array}{ccccc} 3^* & 2 & 4 & 7 \\ 2 & 1 & 1 & 1 \\ 1 & 3 & 5 & 2 \end{array} \right) \xrightarrow{ \frac{1}{2} R_1 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 2 & 1 & 1 & 1 \\ 1 & 3 & 5 & 2 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 2 & 1 & 1 & 1 \\ 1 & 3 & 5 & 2 \end{array} \right) \xrightarrow{R_2 - 2R_1 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & -1/3 & -5/3 & -11/3 \\ 1 & 3 & 5 & 2 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & -1/3 & -5/3 & -11/3 \\ 1 & 3 & 5 & 2 \end{array} \right) \xrightarrow{R_3 - R_1 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & -1/3 & -5/3 & -11/3 \\ 0 & 7/3 & 11/3 & -1/3 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & -1/3 & -5/3 & -11/3 \\ 0 & 7/3 & 11/3 & -1/3 \end{array} \right) \xrightarrow{-3R_2 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & 1 & 5 & 11 \\ 0 & 7/3 & 11/3 & -1/3 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & 1 & 5 & 11 \\ 0 & 7/3 & 11/3 & -1/3 \end{array} \right) \xrightarrow{R_3 - \frac{7}{3} R_2 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & 1 & 5 & 11 \\ 0 & 0 & -8 & -26 \end{array} \right)$$ $$\left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & 1 & 5 & 11 \\ 0 & 0 & -8 & -26 \end{array} \right) \xrightarrow{-\frac{1}{8} R_3 } \left( \begin{array}{ccccc} 1^* & 2/3 & 4/3 & 7/3 \\ 0 & 1 & 5 & 11 \\ 0 & 0 & 1 & 13/4 \end{array} \right)$$ $$\begin{array}{l} z = \frac{13}{4} \\ y = 11 - 5z = 11 - 5\left(\frac{13}{4}\right) = -\frac{21}{4} \\ x = \frac{7}{3} - \frac{2}{3}y - \frac{4}{3}z = \frac{7}{3} - \frac{2}{3}\left(-\frac{21}{4}\right) - \frac{4}{3}\left(\frac{13}{4}\right) = \frac{3}{2} \end{array}$$

Answer:

$$(x, y, z) = \left(\frac {3}{2}, - \frac {21}{4}, \frac {13}{4}\right).$$

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