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Solution to complicated systems of three variables using matrices

Hyunqul

New member
Feb 24, 2012
1
Hello,i have been trying to self-study matrices topics, during that I came across two complicated problems, and I wish I could provided with help to solve them : The question asks to solve each of the following system of equations, using row reduction method (Again...I assure that my teacher has not taught us matrices yet...It's a self attempt) :

First system :
CodeCogsEqn.gif

CodeCogsEqn (1).gif

CodeCogsEqn (2).gif


Second system :

x+(1+pi)y+(1+2pi)z=1+3pi


CodeCogsEqn (3).gif

(1/4)x+(1/2)y+(3/4)z=1


Thanks.
 

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Last edited:

SammyS

New member
Feb 15, 2012
3
Hello,i have been trying to self-study matrices topics, during that I came across two complicated problems, and I wish I could provided with help to solve them : The question asks to solve each of the following system of equations, using row reduction method (Again...I assure that my teacher has not taught us matrices yet...It's a self attempt) :

First system :
View attachment 57

View attachment 61

View attachment 58


Second system :

x+(1+pi)y+(1+2pi)z=1+3pi


View attachment 59

(1/4)x+(1/2)y+(3/4)z=1


Thanks.
For the first system, I assume the first equation should be

$(\log4)x+(3+2\log4)y+(6+3\log4)z=4+9\log4 $



For the second system, divide the second equation by e2. Multiply the last equation by 4.

Both systems have weird coefficients, but are otherwise fairly standard.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The matrix form for the firsrt would be
$\begin{bmatrix}log 4 & 3+ 2log 4 & 6+ 3log 4 \\ 1+ log 3 & 5+ 3log 3 & 3+ 5log 3 \\ -1 & 0 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}9+ 4log 4 \\ 4+ 7log 3 \\ 2\end{bmatrix}$

For the second
$\begin{bmatrix}1 & 1+ \pi & 1+ 2\pi \\ 4e^2 & 3e^3 & 2e^2 \\ \frac{1}{4} & \frac{1}{2} & \frac{3}{4}\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1+ 3\pi \\ e^2 \\ 1\end{bmatrix}$
Now use whatever matrix methods you know, Gauss elimination, inverting, LU decomposition. The numbers are peculiarly written, but they are just numbers.

(Not matrix related but an obvious first step is to divide through the second equation in the second set by $e^2$.)




 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Here's an idea. Re-write your equations as

$\log 4\left(x + 2y + 3z-4\right) + 3(y+2z-3) = 0$

$\left(x+2y+3z-4\right) + \ln 3 \left(x+3y+5z-7\right) = 0$

$x - z + 2 = 0$

If you let

$u =x + 2y + 3z-4$, $v = y+2z-3$ and $w =x+3y+5z-7$

then your system becomes

$\begin{align}
\log 4 u + 3v &= 0,\\
u + \ln 3 w &= 0,\\
u - 2v &= 0,
\end{align}$

whose solution is

$u=0$, $v = 0$ and $w=0$.

Thus, you are required to solve

$\begin{align}
x + 2y + 3z&=4,\\
y+2z&=3,\\
x+3y+5z&=7.
\end{align}$