75032 Matrisemetoder 1996–08–13: Fasit

1a

A = LU = [\begin{matrix}1 & 0 & 0 & 0 \\ a & 1 & 0 & 0 \\ 0 & a & 1 & 0 \\ 1 & 0 & a & 1 \end{matrix}] · [\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & a \\ 0 & 0 & 0 & 0 \end{matrix}]

1b En basis for 𝒩 (A) består av den ene vektoren

[\begin{matrix}−a + 2 \\ 2a − 3 \\ −a \\ 1 \end{matrix}]

mens ℛ(A) har en basis bestående av de første 3 søylene i A. 1c

a = 0 eller a = \frac{1}{2} ± \frac{i}{2}\sqrt{3}.

2c

A = QR = \frac{1}{2} [\begin{matrix}1 & 1 & 1 \\ 1 & −1 & 1 \\ −1 & 1 & 1 \\ −1 & −1 & 1 \end{matrix}] · [\begin{matrix}2 & 2u & 0 \\ 0 & 4 & 2v \\ 0 & 0 & 2 \end{matrix}] .

2d

x = [\begin{matrix}2 − u + uv \\ 1 − v \\ 2 \end{matrix}] .

3

A = QΛQ^{T} = [\begin{matrix}\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} + \frac{λ}{6} & \frac{1}{\sqrt{3}} − \frac{1}{\sqrt{2}} + \frac{λ}{6} & \frac{1}{\sqrt{3}} − \frac{λ}{3} \\ \frac{1}{\sqrt{3}} − \frac{1}{\sqrt{2}} + \frac{λ}{6} & \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} + \frac{λ}{6} & \frac{1}{\sqrt{3}} − \frac{λ}{3} \\ \frac{1}{\sqrt{3}} − \frac{λ}{3} & \frac{1}{\sqrt{3}} − \frac{λ}{3} & \frac{1}{\sqrt{3}} + \frac{2λ}{3} \end{matrix}]