# Consider the following system of equations: x-8y+2z = 4 0x+y+6z = -7 0x+0y+z = 7 Now find the inverse A^-1 of the coefficient matrix A and use it to solve for X.

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You need to build the matrix A, using the coefficients of variables, x,y,z, such that:

`A = ((1,-8,2),(0,1,6),(0,0,1))`

You need to build the column matrix of variables x,y,z, such that:

`X = ((x),(y),(z))`

You need to build the column matrix of constant terms, such that:

`B = ((4),(-7),(7))`

You may write the system as a multiplication of matrices `A,X,B` , such that:

`A*X = B`

You need to multiplicate by `A^(-1)` to the left side of each side of equality, such that:

`A^(-1)*A*X = A^(-1)*B`

Since `A^(-1)*A = I_3` yields:

`I_3*X = A^(-1)*B`

Since `I_3*X = X*I_3= X` yields:

`X = A^(-1)*B`

You need to determine `A^(-1)` using the following formula, such that:

`A^(-1) = (1/(Delta_A))*A^T`

`A^T = ((A_(11),A_(21),A_(31 )),(A_(12),A_(22),A_(32)),(A_(13),A_(23),A_(33)))`

`Delta_A = [(1,-8,2),(0,1,6),(0,0,1)]` => `Delta_A = 1` `=>` `A^(-1) = A^T`

`A_(11) = (-1)^(1+1)*[(1,6),(0,1)] ` => `A_(11) = 1, A_(21) = 8, A_(31) = -50, A_(12) = 0`

`A_(22) = (-1)^(2+2)*[(1,2),(0,1)] ` =>` A_(22) = 1, A_(23) = 0`

`A_(32) = -6, A_(13) = 0, A_(23) = 0 A_(33) = 1`

`A^(-1) = ((1,8,-50),(0,1,-6),(0,0,1))`

You need to evaluate the coefficients x,y,z, solving the matrix multiplication `A^(-1)*B` such that:

`((x),(y),(z)) =` ` ((1,8,-50),(0,1,-6),(0,0,1))`*`((4),(-7),(7))`

`((x),(y),(z)) = ((1*4-7*8-50*7),(0*4-7*1-6*7),(0*4-7*0+1*7))`

`((x),(y),(z)) = ((-402),(-49),(7))`

Hence, evaluating the solution to the system of equations, using `A^(-1),` yields `((x),(y),(z)) = ((-402),(-49),(7)).`

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