 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/18/c/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 18 / c / Solution by

Codex  0 2026-10-18

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients are

b \cdot e_{0} = 0, b \cdot e_{4} = 22, b \cdot e_{6} = 3 .

Thus

b = 22 e_{4} + 3 e_{6}

. If

x = c_{0} e_{0} + c_{4} e_{4} + c_{6} e_{6}

, then matrix multiplication gives

A x = 4 c_{4} e_{4} + 6 c_{6} e_{6}

. Matching coefficients fixes

c_{4} = 2 /2

and

c_{6} = 3 /6

, while

c_{0}

is arbitrary. Consequently all solutions are

x = 2/3 - 1/3 1/6 + t 11 - 2, t \in R .

Geometrically, the solution set is an affine line in a vector space parallel to the kernel of

A

. The zero eigenvalue makes the component in that direction invisible to

A

; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

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