 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014/4/solution

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

gives

100 4 a + 12 0 20 b .

The nonzero rows here have distinct pivots. Therefore the matrix rank is

rank A = ⎩ ⎨ ⎧ 3, 2, 1, a \neq = - 12, b \neq = 0, exactly one of a = - 12, b = 0 holds, a = - 12, b = 0.

There is always at least one nonzero row, so rank zero is impossible.

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