Question

gauss Jordan method 2x+y+z=10 3x+2y+3z=18 x+4y+9z=16​

Answer 1

The solution of the systems of linear equation by Gauss Jordan method is $\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}7\\-9\\5\end{array}\right]$

Step-by-step explanation:

Given: System of linear equations

$2x+y+z=10\\3x+2y+3z=18\\x+4y+9z=16$

To Find: Solution of the given system of linear equations

Solution:

• Finding the values of x, y, and z by Gauss Jordan Method

The system of linear equations can be written as AX = B,

$\left[\begin{array}{ccc}2&1&1\\3&2&3\\1&4&9\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}10\\18\\16\end{array}\right]$

Using row operation $R_1 \rightarrow R_1 = R_1/2$

$\left[\begin{array}{ccc}1&1/2&1/2\\3&2&3\\1&4&9\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}5\\18\\16\end{array}\right]$

Using row operation $R_2 \rightarrow R_2 - 3 R_1$

$\left[\begin{array}{ccc}1&1/2&1/2\\0&1/2&3/2\\1&4&9\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}5\\3\\16\end{array}\right]$

Using row operation $R_3 \rightarrow R_3 - R_1$

$\left[\begin{array}{ccc}1&1/2&1/2\\0&1/2&3/2\\0&7/2&17/2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}5\\3\\11\end{array}\right]$

Using row operation $R_2 \rightarrow 2 R_2$

$\left[\begin{array}{ccc}1&1/2&1/2\\0&1&3\\0&7/2&17/2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}5\\6\\11\end{array}\right]$

Using row operation $R_1 \rightarrow R_1 -R_2/2$

$\left[\begin{array}{ccc}1&0&-1\\0&1&3\\0&7/2&17/2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}2\\6\\11\end{array}\right]$

Using row operation $R_3 \rightarrow R_3 - \frac{7}{2} R_2$

$\left[\begin{array}{ccc}1&0&-1\\0&1&3\\0&0&-2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}2\\6\\-10\end{array}\right]$

Using row operation $R_3 \rightarrow R_3 / (-2)$

$\left[\begin{array}{ccc}1&0&-1\\0&1&3\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}2\\6\\5\end{array}\right]$

Using row operation $R_1 \rightarrow R_1+ R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&3\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}7\\6\\5\end{array}\right]$

Using row operation $R_2 \rightarrow R_2 - 3 R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}7\\-9\\5\end{array}\right]$

Therefore,

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}7\\-9\\5\end{array}\right]$

Hence, the solution of the systems of linear equation by Gauss Jordan method is $\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}7\\-9\\5\end{array}\right]$