 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-i/1/a/iii/solution

R_{2} \leftarrow R_{2} + 2 R_{1}

and

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

, then

R_{2} \leftarrow R_{2} /2

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

, and finally

R_{1} \leftarrow R_{1} + R_{3}

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

1 - 2 1 101223 ⟶ 100120261 ⟶ 100010 - 1 31 ⟶ I_{3} .

I_{3}

. Applying the same operations to the augmented column

b = (b_{1}, b_{2}, b_{3})^{T}

gives

v = b_{3} - b_{1} - \frac{1}{2} b_{2} 4 b_{1} + \frac{1}{2} b_{2} - 3 b_{3} b_{3} - b_{1} .

Thus every

b \in R^{3}

admits exactly one solution. This follows because the elementary row operations preserve the solution set of the augmented linear system, and the reduced coefficient matrix is the identity matrix.

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