Math Help - matrices

1. matrices

Solve x + 10y – 2z = 20 5x – 3y + 4z = 8 2x + y = 6 using matrices thanks

2. Re: matrices

Originally Posted by dzomberg
Solve x + 10y – 2z = 20 5x – 3y + 4z = 8 2x + y = 6 using matrices thanks
You need to find the inverse (if it exists) of
$\left[ {\begin{array}{{rrr}}1&{10}&{ - 2}\\5&{ - 3}&4\\2&1&0\end{array}} \right]$

3. Re: matrices

Hello, dzomberg!

$\text{Solve: }\:\begin{array}{ccc}x + 10y – 2z &=& 20 \\ 5x – 3y + 4z &=& 8 \\ 2x + y\qquad\;\; &=& 6\end{array}\;\;\text{ using matrices}$

$\text{We have: }\:\left[\begin{array}{ccc|c} 1&10 &\text{-}2 & 20 \\ 5&\text{-}3&4&8 \\ 2&1&0&6 \end{array}\right]$

$\begin{array}{c}\\R_2-5R_1 \\ R_3-2R_1 \end{array}\:\left[\begin{array}{ccc|c} 1&10 & \text{-}2 & 20 \\ 0&\text{-}53&14 & \text{-}92 \\ 0 & \text{-}19 & 4 & \text{-}34 \end{array}\right]$

. . $\begin{array}{c}\\\text{-}\frac{1}{53}R_2 \\ \\ \end{array}\:\left[\begin{array}{ccc|c} 1&10&\text{-}2&20 \\ 0&1&\text{-}\frac{14}{53} & \frac{92}{53} \\ 0 & \text{-}19& 4 & \text{-}34\end{array}\right]$

$\begin{array}{c}R_1-10R_2 \\ \\ R_3+19R_2 \end{array}\:\left[\begin{array}{ccc|c} 1&0&\frac{34}{53} & \frac{140}{53} \\ 0&1&\text{-}\frac{14}{53} & \frac{92}{53} \\ 0&0&\text{-}\frac{54}{53} & \text{-}\frac{54}{53} \end{array}\right]$

. . . $\begin{array}{c}\\ \\ \text{-}\frac{53}{54}R_3\end{array}\:\left[\begin{array}{ccc|c} 1&0& \frac{34}{53} & \frac{140}{53} \\ 0&1&\text{-}\frac{14}{53} & \frac{92}{53} \\ 0&0&1&1 \end{array}\right]$

$\begin{array}{c}R_1-\frac{34}{53}R_3 \\ R_2 + \frac{14}{53}R_3 \\ \end{array}\: \left[\begin{array}{ccc|c}1&0&0&2 \\ 0&1&0&2 \\ 0&0&1&1 \end{array}\right]$

$\text{Therefore: }\:\begin{Bmatrix}x &=& 2 \\ y &=& 2 \\ z &=& 1 \end{Bmatrix}$

4. Re: matrices

Originally Posted by dzomberg
Solve x + 10y – 2z = 20 5x – 3y + 4z = 8 2x + y = 6 using matrices thanks

If all you wanted a solution, why not look at this page?