 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/13/b/ii/solution

A^{- 1} = \frac{1}{2} 1 - 7 5 11 - 1 - 2 4 - 2 .

Multiplication by the right-side column yields

r = A^{- 1} d = \frac{1}{2} 1 + 2 - 6 - 7 + 2 + 12 5 - 2 - 6, (x, y, z) = (- \frac{3}{2}, \frac{7}{2}, - \frac{3}{2}) .

Multiplying the inverse by

A

checks the identity matrix, and substitution of this solution into the three original rows gives

1, 2, 3

respectively.

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