How do you write the augmented matrix for the system of linear equations 7x-5y+z=13, 19x=8z=10?

Feb 20, 2017

See explanation

Explanation:

Given:

$7 x - 5 y + z = 13 \text{ } \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$
$19 x = 8 z = 10 \text{ } \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
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Equation 2 can be 'split' into two equations

$7 x - 5 y + z = 13 \text{ } \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$
$19 x \textcolor{w h i t e}{- 5 y + z} = 10 \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left({2}_{a}\right)$
$\textcolor{w h i t e}{7 x - 5 y} + 8 z = 10 \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left({2}_{b}\right)$
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Now this is how we write the augmented matrix

$\left[\begin{matrix}7 & - 5 & 1 & | & 13 \\ 19 & 0 & 0 & | & 10 \\ 0 & 0 & 8 & | & 10\end{matrix}\right]$
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$\textcolor{b l u e}{\text{However; if you meant, how do we solve this. It is as follows}}$

$\left[\begin{matrix}7 & - 5 & 1 & | & 13 \\ 19 & 0 & 0 & | & 10 \\ 0 & 0 & 8 & | & 10\end{matrix}\right]$
${R}_{3} \div 8 \mathmr{and} {R}_{2} \div 19$
$\text{ } \downarrow$

$\left[\begin{matrix}7 & - 5 & 1 & | & 13 \\ 1 & 0 & 0 & | & \frac{10}{19} \\ 0 & 0 & 1 & | & \frac{10}{8}\end{matrix}\right]$
${R}_{1} - 7 {R}_{2}$
$\text{ } \downarrow$

$\left[\begin{matrix}0 & - 5 & 1 & | & \frac{237}{19} \\ 1 & 0 & 0 & | & \frac{10}{19} \\ 0 & 0 & 1 & | & \frac{10}{8}\end{matrix}\right]$
${R}_{1} \div \left(- 5\right)$
$\text{ } \downarrow$

$\left[\begin{matrix}0 & 1 & - \frac{1}{5} & | & - \frac{237}{95} \\ 1 & 0 & 0 & | & \frac{10}{19} \\ 0 & 0 & 1 & | & \frac{10}{8}\end{matrix}\right]$
${R}_{1} + \frac{1}{5} {R}_{3}$
$\text{ } \downarrow$

$\left[\begin{matrix}0 & 1 & 0 & | & - \frac{667}{380} \\ 1 & 0 & 0 & | & \frac{10}{19} \\ 0 & 0 & 1 & | & \frac{10}{8}\end{matrix}\right]$

$\left[\begin{matrix}1 & 0 & 0 & | & \frac{10}{19} \\ 0 & 1 & 0 & | & - \frac{667}{380} \\ 0 & 0 & 1 & | & \frac{5}{4}\end{matrix}\right]$

You will need to check this. It is very easy to go wrong!