Solve the system
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You know the rules by this time!
Please share your work and indicate exactly where you are stuck.
To start of - I'll multiply the second row by (-3) and add that to the first row. What do you get?
Dr.Peterson
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Surely you don't mean that! Please correct it.
These have to be equations, not expressions; and the variables have to be consistent.
But once you get the right equations, start solving from the bottom up.
These have to be equations, not expressions; and the variables have to be consistent.
But once you get the right equations, start solving from the bottom up.
Jagulba
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if you know how to multiply two matrices then always write them down.
Matrix[coefficient]*Matrix[variable] = [constant]
so here would be your matrix (being the coefficient matrix) multiply by the variable matrix
what should be the size of variable matrix (or how many variable are there)? ask yourself if I have a 3*4 matrix could it be multiply by a 2*1 matrix (only x and y as variable)?
Matrix[coefficient]*Matrix[variable] = [constant]
so here would be your matrix (being the coefficient matrix) multiply by the variable matrix
what should be the size of variable matrix (or how many variable are there)? ask yourself if I have a 3*4 matrix could it be multiply by a 2*1 matrix (only x and y as variable)?
Dr.Peterson
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The matrix was
If I call the variables x, y, z for simplicity, the corresponding equations are
Do you see the difference between this and your equations? You have to be careful about where each variable goes.
Now, if your equations had been correct, then (apart from your constantly renaming variables, and using 2 where you meant 5, and doing the arithmetic wrong) your work would have been (sort of) right ...
[MATH]\begin{bmatrix}
1 & -3 & 4 & 7\\
0 & 1 & 2 & 2\\
0 & 0 & 1 & 5
\end{bmatrix}[/MATH]
If I call the variables x, y, z for simplicity, the corresponding equations are
1x - 3y + 4z = 7
0x + 1y + 2z = 2
0x + 0y + 1z = 5
0x + 1y + 2z = 2
0x + 0y + 1z = 5
Do you see the difference between this and your equations? You have to be careful about where each variable goes.
Now, if your equations had been correct, then (apart from your constantly renaming variables, and using 2 where you meant 5, and doing the arithmetic wrong) your work would have been (sort of) right ...
Is the following correct?
1x - 3y + 4z = 7
0x + 1y + 2z = 2
0x + 0y + 1z = 5
x - 3y + 4z = 7 ...(1)
y + 2z = 2 ...(2)
z = 5
plug z into eqn (2)
y + 2(5) = 2
y + 10 = 2
y = 2 - 10
y = -8
plug y into eqn (1)
x - 3(-8) + 4(5) = 7
x + 24 + 20 = 7
x + 44 = 7
x = 7 - 44
x = -37
(-37, -8, 5)
1x - 3y + 4z = 7
0x + 1y + 2z = 2
0x + 0y + 1z = 5
x - 3y + 4z = 7 ...(1)
y + 2z = 2 ...(2)
z = 5
plug z into eqn (2)
y + 2(5) = 2
y + 10 = 2
y = 2 - 10
y = -8
plug y into eqn (1)
x - 3(-8) + 4(5) = 7
x + 24 + 20 = 7
x + 44 = 7
x = 7 - 44
x = -37
(-37, -8, 5)
Jagulba
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It's very interesting and easy way to solve 3 variables and 3 equations.
look here and you get what I mean
the only complex is to make an inverse of your coefficient matrix and simple multiplication and you got the answers
look here and you get what I mean
Solving Systems of Linear Equations Using Matrices
Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.
www.mathsisfun.com
the only complex is to make an inverse of your coefficient matrix and simple multiplication and you got the answers
Dr.Peterson
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The work looks good; have you checked it?
x - 3y + 4z = 7 --> -37 - 3(-8) + 4(5) = 7 good
y + 2z = 2 --> -8 + 2(5) = 2 good
z = 5 good
And now you see what I meant by solving from the bottom up (which is not a standard term to my knowledge). There are more advanced ways to do all this in linear algebra (including what Jagulba has mentioned), but I assume you are just being introduced to matrix methods for solving a system, and have not learned things like matrix multiplication.
x - 3y + 4z = 7 --> -37 - 3(-8) + 4(5) = 7 good
y + 2z = 2 --> -8 + 2(5) = 2 good
z = 5 good
And now you see what I meant by solving from the bottom up (which is not a standard term to my knowledge). There are more advanced ways to do all this in linear algebra (including what Jagulba has mentioned), but I assume you are just being introduced to matrix methods for solving a system, and have not learned things like matrix multiplication.