



 
Hi Karlena, The augmented matrix of a linear system of equations contains the coefficients of the variables augmented by a column for the constants. For your system the augmented matrix is
(I have written the matrix as a table as it is much easier for me to manipulate a table than the standard matrix notation.) The task now is to perform elementary row operations to transform the matrix to have zeros below the diagonal and if preferably with 1 as the diagonal value in each row. (The diagonal of interest starts at the upper left corner so currently the diagonal entries are 1, 3 nd 3.) There are many choices for row operation but here is what I did. Multiply the first row by 2 and add it to the second row. I get
Multiply the first row by 1 and add it to the third row. I get
Multiply the second row by 1 and add it to the third row. I get
Divide the third row by 3. I get
At this point rewrite the system as equations rather than in matrix form.
The last equation tells me z = 8. Substitute z = 8 into the second equation and solve for y. Substitute this value for y and z = 8 into the first equation and solve for x. Verify your answer. Harley  


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