Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 14 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

Expanding the determinant along the first row gives

det A = 1 (- 2 - 1) - 2 (0 + 1) + (- 1) (0 - 1) = - 4.

In particular the matrix is invertible.

ii

Solution

The cofactor matrix is

- 3 5 - 1 - 1 - 1 1 - 1 31

. Transpose it to form the adjugate matrix, and divide by the determinant

- 4

:

A^{- 1} = \frac{1}{4} 311 - 5 1 - 3 1 - 1 - 1 .

Direct matrix multiplication gives

A A^{- 1} = I

, checking all entries.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version