How Do You Solve Systems of Equations Using an Inverse Matrix?

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To solve a system of equations using an inverse matrix, first express the system as a matrix equation Ax=b. The inverse of matrix A can be calculated using the formula A^{-1} = (1/(ad - bc)) * [[d, -b], [-c, a]]. Once the inverse is found, the solution for x can be obtained using x = A^{-1}b. An alternative method involves using row operations to convert the original matrix into the identity matrix, with the inverse appearing on the right side. This approach can be particularly useful for larger matrices and helps reinforce understanding of matrix operations.
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How would I go about solving a systme of equations using an inverse matrix?

i.e. -10x - 6y = 6
7x + y = -7
 
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As I`m sure this is stated in your book (it should be), this system of equations is equivalent to the single matrix equation:

\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)

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.
 
Your question was "How would I go about solving a systme of equations using an inverse matrix?" Are you saying you don't have any problem with actually finding the inverse matrix?
 
Not exactly sure what I am saying. I am not having trouble with any other methods,
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.
 
Still trying to figure this out
 
So can you find the inverse of A?
If:
A:= \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]
Then:
A ^ {-1}= \frac{1}{ad - bc} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right]
After finding the inverse of A, note that:
Ax = b \Leftrightarrow A ^ {-1}Ax = A ^ {-1}b \Leftrightarrow x = A ^ {-1}b
Viet Dao,
 
Although the above method will work (for finding the inverse), you might want to try it a different way. It will help when you get to larger matrices (at least until you learn the generalized form of the above equation).

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

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 A\mathbf{x}=I\implies\mathbf{x}=A^{-1}.
 
SHOW YOUR WORK first.
 
I posted the work I have been doing with Matrices previously, thanks for your concern
 
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