75032 Matrisemetoder 26. mai 1995: Fasit

1a

A = LU = [\begin{matrix}1 \\ 1 & 1 \\ 1 & 2 & 1 \\ 1 & 3 & 3 & 1 \end{matrix}] [\begin{matrix}1 & 1 & 1 & 1 \\ & 1 & 2 & 3 \\ & & 1 & 3 \\ & & & 1 \end{matrix}] = LL^{T}, x = [\begin{matrix}−5 \\ 18 \\ −20 \\ 8 \end{matrix}]

1b Determinanten er 1. 2a Basis for N (A) for eksempel:

For N (A): [\begin{matrix}−2 \\ 1 \\ 0 \\ 0 \end{matrix}] , [\begin{matrix}−7 \\ 0 \\ 1 \\ 1 \end{matrix}] for R(A^{T}): [\begin{matrix}1 \\ 2 \\ 3 \\ 4 \end{matrix}] , [\begin{matrix}0 \\ 0 \\ 1 \\ −1 \end{matrix}] .

2b R(A) = N (A) og N (A^{T}) = R(A^{T}). Bruk basisene fra punkt (a). 2c

J = [\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]

3a

A = Q_{1}ΣQ_{2}^{T}; Q_{1} = [\begin{matrix}2/3 & −1/\sqrt{2} & 1/(3\sqrt{2}) \\ 2/3 & 1/\sqrt{2} & 1/(3\sqrt{2}) \\ 1/3 & 0 & −4/(3\sqrt{2}) \end{matrix}] , Σ = [\begin{matrix}15 & 0 \\ 0 & 5\sqrt{2} \\ 0 & 0 \end{matrix}] , Q_{2} = \frac{1}{5} [\begin{matrix}3 & 4 \\ 4 & −3 \end{matrix}]

3b

x = [\begin{matrix}−1 \\ 1 \end{matrix}] .

4a

[\begin{matrix}1 & 0 & 0 & 0 & … \\ −1 & 1 & 0 & 0 & … \\ 1 & −1 & 1 & 0 & … \\ −1 & 1 & −1 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}] [ \begin{matrix}1 & −1 & 1 & −1 & … \\ 0 & 1 & −1 & 1 & … \\ 0 & 0 & 1 & −1 & … \\ 0 & 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}]

4b

[\begin{matrix}2 & 1 \\ 1 & 2 & 1 \\ & ... & ... & ... \\ & & 1 & 2 & 1 \\ & & & 1 & 1 \end{matrix}] .

5 (1 − r^{2})^{n−1}, |r| < 1.