 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/11/a/i/solution

R_{2} \leftarrow R_{2} - R_{1}

and

R_{3} \leftarrow R_{3} - 2 R_{1}

, then

R_{3} \leftarrow 4 R_{3} - 3 R_{2}

112 2 - 2 1 31 - 1 1401 ⟶ 100 2 - 4 - 3 3 - 2 - 7 14 - 14 - 27 ⟶ 100 2 - 4 0 3 - 2 - 22 14 - 14 - 66 .

By back substitution, the last row gives

z = 3

, the second gives

- 4 y - 6 = - 14

and hence

y = 2

, and the first gives

x + 4 + 9 = 14

. Thus

(x, y, z) = (1, 2, 3) .

The right-hand side has been transformed in every row operation, preserving the solution set.

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