Problem solving idea: Given a system of third-order linear equations, it is as follows: a1xb1yc1z=d1a2xb2yc2z=d2a3xb3yc3z=d3(a1,b1,c1),(a2,b2,c2),(a3,b3,c3) are the row vectors of the coefficient matrix of the system of equations, (d1,d2,d3) are the corresponding constant term vectors, and the goal is to solve the values of variables x, y, z. The steps to solve the problem are as follows: 1. Elementary row transformation: We use a series of row transformations (including swapping two rows, multiplying a row by a non-zero constant, and adding a certain row to several times of another row) to transform the coefficient matrix into a stepped matrix or a simpler row stepped matrix (i.e., an upper triangle matrix). You can try to eliminate the elements below the first column first, and if a1≠0, you can do this by adding the first column.
2. The third line is achieved by subtracting the first row of the appropriate multiple; Otherwise, select a different non-zero row as the first row for the operation. Suppose the following form is obtained after transformation: a1xb1yc1z=d10b2'yc2'z=d2'00c3''z=d3''2. Retrogressive solving: Retrogressive solution starts from an equation with non-zero coefficients. Since the system of equations has been changed to the upper triangular form, the value of z can be calculated directly, i.e., z=d3''/c3''。3. Continue to regression: substitute the obtained z value into the second equation to solve y, that is, y=(d2'-c2'z)/b2'。4. Then solve: substitute the values of y and z into the first equation to solve x, that is, x=(d1-b1y-c1z) a1. 5. Verification results: substitute the obtained x, y, and z values into the original equation system to verify whether all equations are satisfied. The whole process uses the core idea of Gaussian elimination method, simplifies the structure of linear equations through line elementary transformation, and gradually reduces the coupling degree between unknowns, so as to facilitate the solution of the exact values of each variable. This method is suitable for systems of linear equations of any size, but the calculation becomes more complex as the number of equations and unknowns increases. 
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