![mathematica matrix inverse mathematica matrix inverse](https://i.stack.imgur.com/QbDPA.png)
Example Solve the following system of equations using the matrix approach shown above. y −3/2 1 7 Performing the cancellation on the left and the multiplication on the right, we get x −7/2 2 (−7/2)2 + (2)7 7 = y −3/2 1 (−3/2)2 + (1)7 4 and our solution to the system is x = 7, y = 4. Mutiplying the above equation by A−1, we get x −7/2 2 2 −1 A A =. Since the determinant is non zero, the matrix is invertible and 1 7 −4 −7/2 2 −1 A = −3/2 1 −2 3 −2 1 −2 4 −3 7 −2 Recall to find the inverse of the matrix A = −3 Example In our example, we converted the system of equations −2x + 4y = 2 −3x + 7y = 7 to matrix form
![mathematica matrix inverse mathematica matrix inverse](https://cf2.ppt-online.org/files2/slide/r/riQsVhuzlv5jOWFPZJAXH4LM8ed9DyxcwNqInp/slide-53.jpg)
Lets see how this method works in our example. Namely, we can use matrix algebra to multiply both sides of the equation by A−1, thus getting A−1 AX = A−1 B. , we get two equations 7 equating the elements of each matrix, thus getting our linear system back again: Given a system of linear equations in two unknowns −2x + 4y = 2 −3x + 7y = 7 We can solve this system of equations using the matrix identity AX = B, if the matrix A has an inverse. When we identify this matrix with the matrix B =. Using matrix inverses and Mathematica to solve systems of equations (Using 2.4, Goldstein, Schneider and Siegel and Mathematica( available on the OIT website)) Given a system of linear equations in two unknowns −2x + 4y = 2 −3x + 7y = 7 we can write it in matrix form as a single equation AX = B, where −2 4 x 2 A=, X=, B=.