Solving systems of equations by determinants, elimination or substitution

**OzzMan** · October 16th 2007, 05:29 PM

Solve the given sytem of equations

9x+12y+2z=23
21x+2y-5z=-25
-18x+6y-4z=-11

Solve the given system of equations by use of determinants

26L-52M-52N=39
45L+96M+40N=-80
55L+62M-11N=-48

Solve the given system of equation by determinants

6541x+4397y=-7732
3309x-8755y=7622

Solve the given system of equations by either by either substitution or elimination

0.66x+0.66y=-0.77
0.33x+1.32y=1.43

**OzzMan** · October 16th 2007, 08:58 PM

Bump

Anyone know the answers to these?

**topsquark** · October 17th 2007, 03:53 AM

Originally Posted by OzzMan

Bump

Anyone know the answers to these?

Please don't "bump." It's rude and see rule #10 here.

-Dan

**topsquark** · October 17th 2007, 04:05 AM

Originally Posted by OzzMan

Solve the given sytem of equations

9x+12y+2z=23
21x+2y-5z=-25
-18x+6y-4z=-11

Since you have to do a number of these by determinants I'll do this one that way. Specifically, I'm using Cramer's rule.
$\left ( \begin{matrix} 9 & 12 & 2 \\ 21 & 2 & -5 \\ -18 & 6 & -4 \end{matrix} \right ) \left ( \begin{matrix} x \\ y \\ z \end{matrix} \right ) = \left ( \begin{matrix} 23 \\ -25 \\ -11 \end{matrix} \right )$

First we need the determinant of the coefficient matrix. This is 2610.

Now to solve the system. By Cramer's rule we take the vector on the RHS and replace the columns of the coefficient matrix one by one:
$x = \frac{1}{2610} \left | \begin{matrix} 23 & 12 & 2 \\ -25 & 2 & -5 \\ -11 & 6 & -4 \end{matrix} \right | = \frac{-290}{2610} = -\frac{1}{9}$

$y = \frac{1}{2610} \left | \begin{matrix} 9 & 23 & 2 \\ 21 & -25 & -5 \\ -18 & -11 & -4 \end{matrix} \right | = \frac{3045}{2610} = \frac{7}{6}$

$y = \frac{1}{2610} \left | \begin{matrix} 9 & 12 & 23 \\ 21 & 2 & -25 \\ -18 & 6 & -11 \end{matrix} \right | = \frac{13050}{2610} = 5$

-Dan

**earboth** · October 17th 2007, 05:17 AM

Originally Posted by OzzMan

...

Solve the given system of equations by either by either substitution or elimination

0.66x+0.66y=-0.77
0.33x+1.32y=1.43

Hello,

$\left|~\begin{array}{lcr}0.66x+0.66y&=&-0.77 \\0.33x+1.32y&=&1.43\end{array}\right.$ multiply the 2nd equation by (-2):

$\left|~\begin{array}{lcr}0.66x+0.66y&=&-0.77 \\-0.66x-2.64y&=&-2.86\end{array}\right.$ add columnwise(?):

$-1.98y = -3.63~\iff~y = \frac{11}{6}$ plug in this value into the first (or second) equation and solve for x. I've got: $\boxed{x = -3}$

**OzzMan** · October 17th 2007, 03:30 PM

Is there a simple way to evaluate this system of equations without a calculator?

26L-52M-52N=39
45L+96M+40N=-80
55L+62M-11N=-48

**topsquark** · October 17th 2007, 04:25 PM

Originally Posted by OzzMan

Is there a simple way to evaluate this system of equations without a calculator?

26L-52M-52N=39
45L+96M+40N=-80
55L+62M-11N=-48

Simple? No. Without a calculator? Yes. But it's the exact same method and it will take you a while. I'd use Cramer's rule for this as that has less computation than most other methods I know. (Gaussian elimination would be a good alternate as it is easy to check your work.) I would think of this as an excellent way to exercise your algebra skills, but if you are being graded on it, use the calculator!

-Dan

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