Solve Equations using a Matrix

  • Thread starter Thread starter 2slowtogofast
  • Start date Start date
  • Tags Tags
    Matrix
AI Thread Summary
The discussion focuses on solving a system of equations using matrix methods. The original equations are transformed into an augmented matrix, and through a series of row operations, the solution is derived. The final values obtained are x = 1, y = 1, and z = 2, which check out when substituted back into the original equations. Clarifications were made regarding the correct construction of the matrix equation. The user successfully resolves their confusion about the substitution process with assistance from others in the thread.
2slowtogofast
Messages
134
Reaction score
1
solve equations using a matrix

x + y -z = 0
3x - y + z = 4
5x +z = 7

so i wrote the matrix

1 1 -1 0
3 -1 1 4
5 0 1 7

then i multiplied Row 1 by -3 and added that to row 2
and multiplied row 1 by -5 and added that to row 3
resulting in

1 1 -1 0
0 -4 4 4
0 -5 6 7

then i multiplied row 2 by - 1/4

1 1 -1 0
0 1 -1 -1
0 -5 6 7

then multiplied row 2 by 5 and added that to row 3

1 1 -1 0
0 1 -1 -1
0 0 1 2

now i subbed back into eqns

x + y -z = 0
3x - y + z = 4
5x +z = 7

1x + 1y - 1z = 0
0x +1y - 1z = -1
0x + 0y + 1z = 2

so z = 2 and solving all the other equation i get x = 1 and y = 1
is this right because i tried solving these by subtraction and got somthing different
 
Last edited:
Physics news on Phys.org
You have not constructed a matrix equation, you have just rewritten the coefficients in some kind of array/pattern

THe matrix equation is :

\begin{pmatrix}<br /> 1 &amp; 1 &amp; -1\\<br /> 3&amp;-1 &amp; 1\\<br /> 5&amp;0&amp;1<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> x\\<br /> y\\<br /> z<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> 0\\<br /> 4\\<br /> 7<br /> \end{pmatrix}<br /> <br />
 
Last edited:
The augmented matrix that 2slow to go fast used is correct.
All you need to do is plug x y and z back into your original equations and see if they make true statements.
Looks to me like they check out.
CC
 
i just got it i was substituting wrong thanks for the help
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
Back
Top