Just curious

  • Thread starter jeremy222
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  • #1
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An augmented matrix scaled by a number also means the solutions set is scaled by that same number. I believe this is true due to it basically being the same as elementary row operations preformed on each row. Unless it is a zero scalar in which case you lose all conditions. Is my method of thinking about this correct? Its not homework but I think it would help me understand a concept.
 

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  • #2
HallsofIvy
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No, that's not true. If you have an augmented matrix, say
[tex]\begin{bmatrix}a & b & c & j \\ d & e & f & k \\ g & h & i & l\end{bmatrix}[/tex]
That is equivalent to the three equations ax+ by+ cz= j, dx+ ey+ fz= k, gx+ hy+ iz= l.
It is also equivalent to the matrix equation
[tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}j \\ k \\ l\end{bmatrix}[/tex]

"Scaling" the augmented matrix by a number, multiplying every row by that number, gives multiples of the same equations which have the same solutions. If you multiply the coefficient matrix in the second form by a constant, but not the right hand side, that will divide the soluutions by the number.
 
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  • #3
Deveno
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one can look at the 1x1 case:

[a][x] =

if we multiply by 2, let's say, we get:

[2a][x] = [2b].

suppose a ≠ 0. then in either case, we find:

x = b/a, the solution to the 2nd equation is NOT x = 2(b/a).

it is exactly the same for

Ax = b, where A is a matrix, and b is a vector.

if A is invertible, x = A-1b.

and 2Ax = 2b has the same vector solution.

in other words, multiplying an equation on both sides by a number,

doesn't change the solution set for that equation.

if ax + by + cz = d, then for any number m:

m(ax + by + cz) = md, for the same x,y and z.
 

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