Answer to Question #111230 in Linear Algebra for Mzwandile

Question

Given:
A=
4 0 0
2 2 3
1 1 1
B=
4 0 0
2 2 3
1 1 1

Find: a) –A-1+ 3BT b) B-1+ ( AT+A-1)

Answer 1

a) -A^-1+3B^T

$B^T=\begin{pmatrix} 4 & 2 & 1 \\ 0 & 2 & 1 \\ 0 & 3 & 1 \end{pmatrix}$

$A^{-1}=\frac {1}{-4} \begin{pmatrix} -1 & 0 & 0 \\ 1 & 4 & -12 \\ 0 & -4 & 8 \end{pmatrix}=\begin{pmatrix} 0.25 & 0 & 0 \\ -0.25 & -1 & 3 \\ 0 & 1 & -2 \end{pmatrix}$

$-A^{-1}+3B^T=\begin{pmatrix} -0.25 & 0 & 0 \\ 0.25 & 1 & -3 \\ 0 & -1 & 2 \end{pmatrix}+\begin{pmatrix} 12 & 6 & 3 \\ 0 & 6 & 3 \\ 0 & 9 & 3 \end{pmatrix}=\begin{pmatrix} 11.75 & 6 & 3 \\ 0.25 & 7 & 0 \\ 0 & 8 & 5 \end{pmatrix}$

Answer:

$-A^{-1}+3B^T=\begin{pmatrix} 11.75 & 6 & 3 \\ 0.25 & 7 & 0 \\ 0 & 8 & 5 \end{pmatrix}$

b) B^-1+(A^T+A^-1)

$B^{-1}+(A^T+A^{-1})=\begin{pmatrix} 0.25 & 0 & 0 \\ -0.25 & -1 & 3 \\ 0 & 1 & -2 \end{pmatrix}+( \begin{pmatrix} 4 & 2 & 1 \\ 0 & 2 & 1 \\ 0 & 3 & 1 \end{pmatrix}+$

$\begin{pmatrix} 0.25 & 0 & 0 \\ -0.25 & -1 & 3 \\ 0 & 1 & -2 \end{pmatrix})=\begin{pmatrix} 4.5 & 2 & 1 \\ -0.5 & 0 & 7 \\ 0 & 5 & -3 \end{pmatrix}$

Answer:

$B^{-1}+(A^T+A^{-1})=\begin{pmatrix} 4.5 & 2 & 1 \\ -0.5 & 0 & 7 \\ 0 & 5 & -3 \end{pmatrix}$