Augmented matrix

kabdul02 · **Joined:** Mon, 16 Feb 2009 06:30:28 UTC **Posts:** 5

can someone please help me solve this question?
I tried so many times but I still could not get it...

3x + y = 2
4x â€“ 2y = -3

the ans is supposed to be (1/10,17/10))

thanks in advance
much appreciated

aswoods · **Posted:** Tue, 3 Mar 2009 21:04:55 UTC

Multiply the first equation by 4, and multiply the second equation by 3. Then subtract one from the other.

kabdul02 · **Joined:** Mon, 16 Feb 2009 06:30:28 UTC **Posts:** 5

augumentation goes like this

[3 1 | 2]
[4 -2 | -3]

i dunno how to implement the next step?

ans =1/10,17/10

tashirosgt · **Posted:** Wed, 4 Mar 2009 06:24:14 UTC

Let the new second row be the old second row minus (4/3) times the first row.

kabdul02 · **Joined:** Mon, 16 Feb 2009 06:30:28 UTC **Posts:** 5

i still dont get it!

Ilaggoodly · **Joined:** Thu, 15 Feb 2007 06:35:15 UTC **Posts:** 755

$\begin{bmatrix} 3 & 1 & | & 2\\ 4 & -2 & | & -3\\ \end{bmatrix}$

We want to get it into reduced row echelon form, so the first step is to snag a 0 in the lower left corner using only elementary row operations (addition, scalar mulitplication, swap)

so, lets subtract 4/3 row 1 from row 2

$\begin{bmatrix} 3 & 1 & | & 2\\ 0 & -2 - 4/3 & | & -3 - 2*(4/3)\\ \end{bmatrix}$

okay so now its in row echelon form, lets get it to reduced row echelon form by getting a zero in the top right corner (not including the augmentation obv.)

in the end you can mulitply each row by a scalar in order to get

$\begin{bmatrix} 1 & 0 & | & A\\ 0 & 1 & | & B\\ \end{bmatrix}$

where A and B are your answers

Soroban · **Posted:** Sat, 7 Mar 2009 19:28:40 UTC

Hello, kabdul02!

Quote:

$\begin{array}{ccc}3x + y &=& 2 \\ 4x - 2y &=& \text{-}3 \end{array}$ . . Answer: . $\left(\frac{1}{10},\:\frac{17}{10}\right)$

I'll do it head-on . . .

$\begin{array}{c} \\ R_2-R_1\end{array} \left|\begin{array}{cc|c} 3 & 1 & 2 \\ 1 & \text{-}3 & \text{-}5 \end{array}\right|$

$\begin{array}{c}R_1 - 2R_2 \\ \\ \end{array} \left|\begin{array}{cc|c} 1 & 7 & 12 \\ 1 & \text{-}3 & \text{-}5 \end{array}\right|$

$\begin{array}{c}\\ R_2-R_1 \end{array} \left|\begin{array}{cc|c}1 & 7 & 12 & 0 & \text{-}10 & \text{-}17 \end{array}\right|$

. $\begin{array}{c}\\ \text{-}\frac{1}{10}R_2 \end{array} \left| \begin{array}{cc|c}1 & 7 & 12 \\ 0 & 1 & \frac{17}{10} \end{array}\right|$

$\begin{array}{c}R_1-7R_2 \\ \\ \end{array} \left|\begin{array}{cc|c} 1 & 0 & \frac{1}{10} \\ \\[-4mm] 0 & 1 & \frac{17}{10} \end{array}\right|$

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Augmented matrix

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