1. ## matrix question 1

.

How to use row-reduce method to solve?

2. $\left[ \begin{array}{ccc|c}1 & 1 & 1 & 4 \\ 9 & 3 & 1 & 8 \\ 4 & -2 & 1 & 8 \end{array}\right]$

$R_2 ' \rightarrow R_2 - 9 R_1$ : $\left[ \begin{array}{ccc|c}1 & 1 & 1 & 4 \\ 0 & -6 & -8 & -28 \\ 4 & -2 & 1 & 8 \end{array}\right]$

$R_3 ' \rightarrow R_3 - 4R_1$ : $\left[ \begin{array}{ccc|c}1 & 1 & 1 & 4 \\ 0 & -6 & -8 & -28 \\ 0 & -6 & -3 & -8 \end{array}\right]$

$R_3 ' \rightarrow R_3 - R_2$ : $\left[ \begin{array}{ccc|c}1 & 1 & 1 & 4 \\ 0 & -6 & -8 & -28 \\ 0 & 0 & 5 & 20 \end{array}\right]$

$\begin{array}{rl} R_1 ' & \rightarrow R_1 + \frac{1}{6}R_2 \\ R_3 ' & \rightarrow \frac{1}{5} R_3 \end{array}$ : $\left[ \begin{array}{ccc|c}1 & 0 & -\frac{1}{3} & -\frac{2}{3} \\ 0 & -6 & -8 & -28 \\ 0 & 0 & 1 & 4 \end{array}\right]$

Try finishing up. The idea is to make all entries 0's in each column besides the leading 1 by using the row containing the leading 1 of that column (you may want to re-read that). Usually starting from the left to the right does the job.

3. Hello, Faz!

Didn't I solve this one already?

$\left[\begin{array}{ccc|c}1&1&1&4 \\ 9&3&1&8 \\ 4&\text{-}2&1&8 \end{array}\right]$
I try to avoid fractions as much as possible . . .

$\begin{array}{c}\\R_2-9R_1 \\ R_3-4R_1\end{array}\left[\begin{array}{ccc|c}1&1&1&4 \\ 0&\text{-}6&\text{-}8&\text{-}28\\0&\text{-}6&\text{-}3&\text{-}8 \end{array}\right]$

$\begin{array}{c}\\ \text{-}\frac{1}{2}R_2 \\ R_3-R_2\end{array} \left[\begin{array}{ccc|c}1&1&1&4 \\ 0&3&4&14 \\ 0&0&5&20 \end{array}\right]$

. . . $\begin{array}{c}\\ \\ \frac{1}{5}R_3\end{array} \left[\begin{array}{ccc|c}1&1&1&4 \\ 0&3&4&14 \\ 0&0&1&4 \end{array}\right]$

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

. . . $\begin{array}{c}\\ \frac{1}{3}R_2 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&1&0&0 \\ 0&1&0&\text{-}\frac{2}{3} \\ 0&0&1&4 \end{array}\right]$

$\begin{array}{c}R_1-R_2 \\ \\ \\ \end{array} \left[ \begin{array}{ccc|c}1&0&0&\frac{2}{3} \\ \\[-4mm] 0&1&0&\text{-}\frac{2}{3} \\ \\[-4mm] 0&0&1&4 \end{array}\right]$

Therefore: . $\begin{Bmatrix}x &=& \frac{2}{3} \\ \\[-3mm] y &=& \text{-}\frac{2}{3} \\ \\[-3mm] z&=&4 \end{Bmatrix}$