Systems of Linear Equations

**chrozer** · April 17th 2009, 09:25 PM

Solve the system of linear equations algebraically.

$x_1 - x_2 + 2x_3 + 2x_4 + 6x_5 = 6$

$3x_1 - 2x_2 + 4x_3 + 4x_4 + 12x_5 = 14$

$-x_2 - x_3 - x_4 - 3x_5 = -3$

$2x_1 - 2x_2 + 4x_3 + 5x_4 + 15x_5 = 10$

$2x_1 - 2x_2 + 4x_3 + 4x_4 + 13x_5 = 13$

I tried to do this many times using various different ways, but I could never get the answer. Help?

I also attached a picture of the original problem for reference.

**Soroban** · April 18th 2009, 02:23 AM

Hello, chrozer!

I used an Augmented Matrix . . . it took me two tries . . .

Solve the system of linear equations algebraically.

. . $\begin{array}{ccc}x_1 - x_2 + 2x_3 + 2x_4 + 6x_5 &=& 6 \\ 3x_1 - 2x_2 + 4x_3 + 4x_4 + 12x_5 &=& 14 \\ \qquad -x_2 - x_3 - x_4 - 3x_5 &=& \text{-}3 \\ 2x_1 - 2x_2 + 4x_3 + 5x_4 + 15x_5 &=& 10 \\ 2x_1 - 2x_2 + 4x_3 + 4x_4 + 13x_5 &=& 13\end{array}$

Multiply the third equation by -1 . . .

We have: . $\left[ \begin{array}{ccccc|c} 1&\text{-}1&2&2&6&6 \\ 3&\text{-}2&4&4&12&14 \\ 0&1&1&1&3&3 \\ 2&\text{-}2&4&5&15&10 \\ 2&\text{-}2&4&4&13&13 \end{array}\right]$

$\begin{array}{c}\\ R_2-3R_1 \\ \\ R_4-2R_1 \\ R_5-R_4 \end{array} \left[ \begin{array}{ccccc|c} 1&\text{-}1&2&2&6&6 \\ 0&1&\text{-}2&\text{-}2&\text{-}6&\text{-}4 \\ 0&1&1&1&3&3 \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&\text{-}1&\text{-}2&3 \end{array}\right]$

$\begin{array}{c}R_1 +R_2 \\ \\ R_3-R_2 \\ \\ R_5+R_4 \end{array} \left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&\text{-}1&\text{-}2&\text{-}6&\text{-}4 \\ 0&0&3&3&9&7 \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&0&1&5 \end{array}\right]$

. . . $\begin{array}{c} \\ \\ \frac{1}{2}R_3 \\ \\ \\ \end{array} \left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&\text{-}2&\text{-}2&\text{-}6&\text{-}4 \\ 0&0&1&1&3&\frac{7}{3} \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&0&1 & 1 \end{array}\right]$

$\begin{array}{c} \\ R_2+2R_3 \\ R_3 - R_4 \\ R_4 - 3R_5 \\ \\ \end{array}\left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&0&0&0&\frac{2}{3} \\ \\[-4mm] 0&0&1&0&0&\frac{13}{3} \\ 0&0&0&1&0& \text{-}5 \\ 0&0&0&0&1&1 \end{array}\right]$

**stapel** · April 18th 2009, 07:18 AM

Originally Posted by chrozer

[B]I tried to do this many times using various different ways, but I could never get the answer.

Since we cannot see your steps, it will not be possible to find any errors. Sorry!

What methods are you allowed to use? For instance, can you plug this into your graphing calculator? Can you use Cramer's Rule, with your calculator computing the determinants?

Thank you!

**chrozer** · April 18th 2009, 08:06 AM

Originally Posted by stapel

Since we cannot see your steps, it will not be possible to find any errors. Sorry!

What methods are you allowed to use? For instance, can you plug this into your graphing calculator? Can you use Cramer's Rule, with your calculator computing the determinants?

Thank you!

I wish we could use a graphing calculator, it would have made it way easier by using a matrix.

I think we have to solve it algebraically through elimination or substitution. That's what I tried to do as that was the way our teacher thought us.

**Soroban** · April 18th 2009, 12:45 PM

Hello, chrozer!

We can do it by Elimination . . . but it takes longer and is quite messy.

I hate those subscripts; I'll change the variables.
And I'll multiply the third equation by -1.

Solve the system of linear equations algebraically.

$\begin{array}{cccc}a - b + 2c + 2d + 6e &=& 6 & [E1]\\ 3a - 2b + 4c + 4d + 12e &=& 14 &[E2]\\ \qquad b + \;c + \;d + \;3e &=& 3 &[E3]\\ 2a - 2b + 4c + 5d + 15e &=& 10 &[E4]\\ 2a - 2b + 4c + 4d + 13e &=& 13 &[E5]\end{array}$

$\begin{array}{ccccc}\text{-}2(E1)\!: & \text{-}2a + 2b - 4c - 4d - 12e &=&\text{-}12 \\ \text{add }(E4)\!: & 2a - 2b + 4c + 5d + 15e &=& 10 \end{array}$

. . and we have: . $d + 3e \:=\:\text{-}2\;\;[E6]$

$\begin{array}{ccccc}\text{-}1(E4)\!: & \text{-}2a + 2b - 4c - 5d - 15e &=& \text{-}10 \\ \text{add }(E5)\!: & 2a - 2b + 4c + 4c + 13e &=& 13 \end{array}$

. . and we have: . $\text{-}d-2e \:=\:3\;\;[E7]$

Add [E6] and [E7]: . $\boxed{e \:=\:1}$

Substitute into [E6]: . $d + 3(1) \:=\:\text{-}2 \quad\Rightarrow\quad\boxed{d \:=\:\text{-}5}$

Substitute into [E1], [E2], [E3]:

. . $\begin{array}{cccccccc}a-b+2c + 2(\text{-}5) + 6(1) &=& 6 & \Longrightarrow & a - b + 2c &=&10 & [E8] \\ 3a-2b+4c + 4(\text{-}5) + 12(1) &=&14 & \Longrightarrow & 3a - 2b + 4c &=& 22 & [E9] \\ \qquad b+c+(\text{-}5) + 3(1) &=& 3 & \Longrightarrow & \qquad b + c &=& 5 & [E10]\end{array}$

$\begin{array}{cccc}-3(E8)\!:& \text{-}3a + 3b - 6c &=& -30 \\ \text{add }[E9]\!: & 3a-2b + 4c &=& 22 \end{array}$

$\begin{array}{ccccc}\text{We have:} &b - 2c &=& \text{-}8 \\ \text{subtract }(E10)\!: & \text{-}b - c &=& \text{-}5 \end{array}$

And we have: . $\text{-}3c \:=\:\text{-}13 \quad\Rightarrow\quad\boxed{c \:=\:\tfrac{13}{3}}$

Substitute into (E10): . $b + \tfrac{13}{3} \:=\:5 \quad\Rightarrow\quad\boxed{b \:=\:\tfrac{2}{3}}$

Substitute into (E8): . $a - \tfrac{2}{3} + 2\left(\tfrac{13}{3}\right) \:=\:10 \quad\Rightarrow\quad\boxed{a \:=\:2}$

Therefore: . $a \:=\: 2,\quad b\:=\:\tfrac{2}{3},\quad c \:=\: \tfrac{13}{3},\quad d \:=\:\text{-}5, \quad e \:=\: 1$

**chrozer** · April 18th 2009, 01:23 PM

Ok I see what I did wrong now. Thanks for the help.

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