- #1

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i.e. -10x - 6y = 6

7x + y = -7

- Thread starter TonyC
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- #1

- 86

- 0

i.e. -10x - 6y = 6

7x + y = -7

- #2

Galileo

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[tex]\left( \begin{array}{cc} -10 & -6 \\ 7 & 1\end{array}\right)\left( \begin{array}{c} x \\ y \end{array}\right)=\left( \begin{array}{c} 6 \\ -7 \end{array}\right)[/tex]

In general, you can write a linear system of n equations in n unknowns as a matrix equation Ax=b. If you can find the inverse of A (let's call it C, so that CA=AC=I) you have solved the problem, just put x=Cb, then A(Cb)=(AC)b=b.

- #3

HallsofIvy

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- #4

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i.e. Cramer Method, Gaussian, elementary row operations. This one has me truly stumped. Maybe I'm too tied (or old) to grasp it today.

- #5

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Still trying to figure this out

- #6

VietDao29

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If:

[tex]A:= \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right][/tex]

Then:

[tex]A ^ {-1}= \frac{1}{ad - bc} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right][/tex]

After finding the inverse of A, note that:

[tex]Ax = b \Leftrightarrow A ^ {-1}Ax = A ^ {-1}b \Leftrightarrow x = A ^ {-1}b[/tex]

Viet Dao,

- #7

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[tex]\left[\begin{array}{cc|cc}a & b & 1 & 0 \\ c & d & 0 & 1\end{array}\right][/tex]

Use row operations to make the left side (your original matrix) the identity matrix, and what is left on the right will be the inverse. You are basically saying [itex]A\mathbf{x}=I\implies\mathbf{x}=A^{-1}[/itex].

- #8

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SHOW YOUR WORK first.

- #9

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I posted the work I have been doing with Matrices previously, thanks for your concern

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