Question

solve the equations by matrix method x+2y-3z=6,2x-y+4z=2,2x+3y-2z=14​

Answer 1

We have the matrix,

$\left[\begin{array}{ccc|c}1&2&-3&6\\2&-1&4&2\\2&3&-2&14\end{array}\right]$

Let's begin!

$\left[\begin{array}{ccc|c}1&2&-3&6\\2&-1&4&2\\2&3&-2&14\end{array}\right]\xrightarrow{R_3-R_2\to R_2}\left[\begin{array}{ccc|c}1&2&-3&6\\0&4&-6&12\\2&3&-2&14\end{array}\right]\\\\\\\left[\begin{array}{ccc|c}1&2&-3&6\\0&4&-6&12\\2&3&-2&14\end{array}\right]\xrightarrow{R_3-2R_1\to R_3}\left[\begin{array}{ccc|c}1&2&-3&6\\0&4&-6&12\\0&-1&4&2\end{array}\right]$

$\left[\begin{array}{ccc|c}1&2&-3&6\\0&4&-6&12\\0&-1&4&2\end{array}\right]\xrightarrow{R_1-\frac{1}{2}R_2\to R_1}\left[\begin{array}{ccc|c}1&0&0&0\\0&4&-6&12\\0&-1&4&2\end{array}\right]\\\\\\\left[\begin{array}{ccc|c}1&0&0&0\\0&4&-6&12\\0&-1&4&2\end{array}\right]\xrightarrow{R_2+3R_3\to R_2}\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&-1&4&2\end{array}\right]$

$\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&-1&4&2\end{array}\right]\xrightarrow{R_2+R_3\to R_3}\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&0&10&20\end{array}\right]\\\\\\\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&0&10&20\end{array}\right]\xrightarrow{\frac{1}{10}R_3\to R_3}\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&0&1&2\end{array}\right]$

$\left[\begin{array}{ccc|c}1&0&0&0\\0&1&6&18\\0&0&1&2\end{array}\right]\xrightarrow{R_2-6R_3\to R_2}\left[\begin{array}{ccc|c}1&0&0&0\\0&1&0&6\\0&0&1&2\end{array}\right]$

Finally,

$\mathbf{x=0\quad;\quad y=6\quad;\quad z=2}$