Question

Solve by gauss Jordan method
2x +9y + 3z= 1
x+ 5y+ 7z= 2
3x +2y+11z = 3​

Answer 1

$\huge\underline{\sf{Solution-}}$

Let consider the equation in the following order,

• x + 5y + 7z = 2

• 2x + 9y + 3z = 1

• 3x + 2y + 11z = 3

The matrix form of the above equation is

$\rm :\longmapsto\:\begin{gathered}\sf A=\left[\begin{array}{ccc}1&5&7\\2&9&3\\3&2&11\end{array}\right]\end{gathered}, \: \: \begin{gathered}\sf B=\left[\begin{array}{c}2\\1\\3\end{array}\right]\end{gathered}$

The augmented matrix is given by

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&5&7&2\\2&9&3&1\\3&2&11&3\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \to R_2 - 2R_1$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&5&7&2\\0& - 1& - 11& - 3\\3&2&11&3\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_3 \to R_3 - 3R_1$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&5&7&2\\0& - 1& - 11& - 3\\0& - 13& - 10& - 3\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \to - R_2 \\ \rm :\longmapsto\:OP \: R_3 \to - R_3$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&5&7&2\\0& 1& 11& 3\\0& 13& 10& 3\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_3 \to R_3 - 13R_2$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&5&7&2\\0& 1& 11& 3\\0& 0& - 133& - 36\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_1 \to R_1 - 5R_2$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&0& - 48& - 13\\0& 1& 11& 3\\0& 0& - 133& - 36\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_3 \to \: - \: \dfrac{1}{133} R_3$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&0& - 48& - 13\\0& 1& 11& 3\\0& 0& 1& \dfrac{36}{133} \end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_2 \to R_2 - 11R_3$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&0& - 48& - 13\\0& 1& 0& \dfrac{3}{133} \\0& 0& 1& \dfrac{36}{133} \end{array}\right]\end{gathered}$

$\rm :\longmapsto\:OP \: R_1 \to R_1 + 48R_3$

$\begin{gathered}\sf [A: B]=\left[\begin{array}{cccc}1&0& 0& - \dfrac{1}{133} \\0& 1& 0& \dfrac{3}{133} \\0& 0& 1& \dfrac{36}{133} \end{array}\right]\end{gathered}$

$\rm :\implies\:x = - \dfrac{1}{133}, \: y = \dfrac{3}{133}, \: z = \dfrac{36}{133}$

Verification : -

Consider the equation

$\rm :\longmapsto\:x + 5y + 7z = 2$

On substituting the values of x, y and z, we get

$\rm :\longmapsto\: - \dfrac{1}{133} + 5 \times \dfrac{3}{133} + 7 \times \dfrac{36}{133} = 2$

$\rm :\longmapsto\: - \dfrac{1}{133} + \dfrac{15}{133} + \dfrac{252}{133} = 2$

$\rm :\longmapsto\:\dfrac{ - 1 + 15 + 252}{133} = 2$

$\rm :\longmapsto\:\dfrac{266}{133} = 2$

$\rm :\longmapsto\:2 = 2$

Hence, Verified

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{x = - \dfrac{1}{133} } \\ \\ &\sf{y = \dfrac{3}{133} }\\ \\ &\sf{z = \dfrac{36}{133} } \end{cases}\end{gathered}\end{gathered}$