Answer to Question #111230 in Linear Algebra for Mzwandile

Question #111230
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)
1
Expert's answer
2020-04-30T15:18:14-0400

a) -A-1+3BT

BT=(421021031)B^T=\begin{pmatrix} 4 & 2 & 1 \\ 0 & 2 & 1 \\ 0 & 3 & 1 \end{pmatrix}

A1=14(1001412048)=(0.25000.2513012)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}

A1+3BT=(0.25000.2513012)+(1263063093)=(11.75630.2570085)-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:

A1+3BT=(11.75630.2570085)-A^{-1}+3B^T=\begin{pmatrix} 11.75 & 6 & 3 \\ 0.25 & 7 & 0 \\ 0 & 8 & 5 \end{pmatrix}


b) B-1+(AT+A-1)

B1+(AT+A1)=(0.25000.2513012)+((421021031)+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}+

(0.25000.2513012))=(4.5210.507053)\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:

B1+(AT+A1)=(4.5210.507053)B^{-1}+(A^T+A^{-1})=\begin{pmatrix} 4.5 & 2 & 1 \\ -0.5 & 0 & 7 \\ 0 & 5 & -3 \end{pmatrix}


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