As Tia solved the system of equations, she transformed matrices at different steps during the process. negative A minus b plus 2 c is equal to 7 2 A plus b plus c is equal to 2 negative 3 A plus 2 b plus 3 c is equal to 7 She noted the following matrices. I II III IV the 3 by 4 matrix, row 1, column 1, 1 column 2, 0 column 3, 0 column 4, 0 row 2, column 1, 0 column 2, 1 column 3, 0 column 4, negative 1 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3 the 3 by 4 matrix, row 1, column 1, negative 1 column 2, negative 1 column 3, 2 column 4, 7 row 2, column 1, 2 column 2, 1 column 3, 1 column 4, 2 row 3, column 1, negative 3 column 2, 2 column 3, 3 column 4, 7 the 3 by 4 matrix, row 1, column 1, 1 column 2, 1 column 3, negative 2 column 4, negative 7 row 2, column 1, 0 column 2, 1 column 3, negative 5 column 4, negative 16 row 3, column 1, 0 column 2, 0 column 3, 22 column 4, 66 the 3 by 4 matrix, row 1, column 1, 1 column 2, 1 column 3, 0 column 4, negative 1 row 2, column 1, 0 column 2, 1 column 3, 0 column 4, negative 1 row 3, column 1, 0 column 2, 0 column 3, 1 column 4, 3 In which order should the matrices be arranged when solving the system from start to finish?
Matrix II is the starting augmented matrix because its rows contain the coefficients and constants from the original equations. Matrix III comes next after row operations create zeros below the first and second pivots. Further elimination gives matrix IV, where the system shows a + b = −1, b = −1, and c = 3. One final row operation changes the first row to a = 0, giving matrix I. So the order is II, III, IV, I.
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