Solve Equations using a Matrix

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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
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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
 
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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 />
 
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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
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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