Solution to complicated systems of three variables using matrices

[Hyunqul]

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 58Second system :

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

View attachment 59

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


Thanks.

 

[SammyS]

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

https://www.physicsforums.com/attachments/58Second system :

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

https://www.physicsforums.com/attachments/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]

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 $e^2$.)

 

[Mathwebster]

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

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

 

[blue_raver22]

Hello,

It is great that you are self-studying matrices and attempting to solve these complicated systems of equations. Using matrices is a powerful tool in solving systems of equations, and the row reduction method is one approach that can be used.

For the first system, we can write the equations in matrix form as:

[1 2 -3 | -2]

[2 -1 1 | 6]

[3 1 -1 | 5]

To solve this system using the row reduction method, we can use elementary row operations to transform the matrix into an upper triangular form. This can be done by multiplying a row by a scalar, adding a multiple of one row to another, or swapping rows. The goal is to get zeros below the diagonal elements.

After performing these operations, we get the following matrix:

[1 2 -3 | -2]

[0 -5 7 | 10]

[0 0 0 | -1]

From this, we can see that the third equation is inconsistent, meaning there is no solution for the system. The first two equations can be rewritten as:

x + 2y - 3z = -2

-5y + 7z = 10

Using back substitution, we can solve for y and z in terms of x. However, since the third equation has no solution, the system has no unique solution.

For the second system, we can write the equations in matrix form as:

[1 1+pi 1+2pi | 1+3pi]

[1/4 1/2 3/4 | 1]

Using the row reduction method, we can perform the same elementary row operations to transform the matrix into an upper triangular form. After performing these operations, we get the following matrix:

[1 1+pi 1+2pi | 1+3pi]

[0 1/4 1/4 | 1+pi]

From this, we can see that the system has a unique solution for x, y, and z. Using back substitution, we can solve for x, y, and z in terms of pi.

I hope this helps you understand how to use matrices to solve systems of equations. Keep up the self-studying and good luck with your future studies!