# Write down the following equations in matrix form. Using inverse of matrix, solve the equations, X + Y + Z = 6 3X – Y + 3Z = 10 5X + 5Y – 4Z = 3

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You need to collect the coefficients of variables x,y,z to form the matrix of the system, such that:

A = ((1,1,1),(3,-1,3),(5,5,-4))

You may evaluate the determinant of matrox of system, such that:

`Delta = [(1,1,1),(3,-1,3),(5,5,-4)] ` => `Delta = 4 + 15 + 15 + 5 - `

`15 + 12 Delta = 36`

Since `Delta != 0` , you may solve the system of equations using Cramer's method, such that:

`x = (Delta_x)/Delta, y = (Delta_y)/Delta, z = (Delta_z)/Delta`

`Delta_x = [(6,1,1),(10,-1,3),(3,5,-4)]`

`Delta_x = 24 + 50 + 9 + 3 - 90 + 40 = 36`

`x = 36/36 => x = 1`

`Delta_y = [(1,6,1),(3,10,3),(5,3,-4)] ` => `Delta_y = -40 + 9 + 90 - 50 - 9 + 72 = 72`

`y = 72/36 => y = 2`

Replacing 1 for x and 2 for y in the top equation `x + y + z = 6 ` yields:

`1 + 2 + z = 6 => z = 3`

**Hence, evaluating the solution to the system of equations, using matrices, yields**` x = 1, y = 2, z = 3.`

The instructions ask us to solve the system using **inverse matrices**:

The matrix form of the equation is:

`([1,1,1],[3,-1,3],[5,5,-4])([x],[y],[z])=([6],[10],[3])`

To solve an equation of the form `AX=B` we find the inverse of A, then `A^(-1)AX=A^(-1)B ==> X=A^(-1)B` making sure to multiply in the same order.

Using technology the inverse is `([-11/36,1/4,1/9],[3/4,-1/4,0],[5/9,0,-1/9])`

Then `X=([x],[y],[z])=([-11/36,1/4,1/9],[3/4,-1/4,0],[5/9,0,-1/9])([6],[10],[3])`

or `X=([1],[2],[3])`

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If you are asked to find the inverse by hand then we use Gauss-Jordan elimination on the augmented matrix:

`([1,1,1:1,0,0],[3,-1,3:0,1,0],[5,5,-4:0,0,1])` until the left side is in reduced row echelon form.