Question #311850

Given

[1 1 -1]

A=[2 2 3]

[4 0 0]

[1 1 -1]

b=[2 2 3]

[4 0 0]

1.Find –A-1+ 3BT

2.Find B-1+ ( AT+A-1)


1
Expert's answer
2022-03-19T02:41:29-0400


Given that


A=[111223400]A=\begin{bmatrix} 1 & 1 & -1\\ 2 & 2 & 3\\ 4 & 0 &0 \end{bmatrix} and B=[111223400]B=\begin{bmatrix} 1 & 1 & -1\\ 2 & 2 & 3\\ 4 & 0 &0 \end{bmatrix}


Then


AT=[124120130]A^{T}=\begin{bmatrix} 1 & 2 & 4\\ 1 & 2 & 0\\ -1 & 3 &0 \end{bmatrix} and BT=[124120130]B^{T}=\begin{bmatrix} 1 & 2 & 4\\ 1 & 2 & 0\\ -1 & 3 &0 \end{bmatrix}


det(A)=111223400det(A)=\begin{vmatrix} 1 & 1 & -1\\ 2 & 2 & 3\\ 4 & 0 &0 \end{vmatrix}


det(A)=(1)(00)(1)(012)+(1)(08)det(A)=(1)(0-0)-(1)(0-12)+(-1)(0-8)


det(A)=0+12+8det(A)=0+12+8


det(A)=20det(A)=20


And


det(B)=111223400det(B)=\begin{vmatrix} 1 & 1 & -1\\ 2 & 2 & 3\\ 4 & 0 &0 \end{vmatrix}


det(B)=(1)(00)(1)(012)+(1)(08)det(B)=(1)(0-0)-(1)(0-12)+(-1)(0-8)


det(B)=0+12+8det(B)=0+12+8


det(B)=20det(B)=20




For the matrix A=[111223400]A=\begin{bmatrix} 1 &1&-1 \\ 2 & 2&3\\ 4&0&0 \end{bmatrix}


A1=adjointAdet(A)A^{-1}=\frac{adjointA}{det(A)}


The adjoint of A is calculated by the cofactor method


A1=[0051245840](20)A^{-1}=\frac{\begin{bmatrix} 0 &0&5 \\ 12&4&-5 \\ -8 & 4&0 \end{bmatrix}}{(20)}


A1=[0200205201220420320820420020]A^{-1}=\begin{bmatrix} \frac{0}{20} &\frac{0}{20}&\frac{5}{20} \\ \frac{12}{20} & \frac{4}{20}&\frac{3}{20}\\ \frac{-8}{20}&\frac{4}{20}&\frac{0}{20} \end{bmatrix}


A1=[000.250.60.20.250.40.20]A^{-1}=\begin{bmatrix} 0 &0 & 0.25\\ 0.6 & 0.2 & -0.25\\ -0.4 & 0.2 &0 \end{bmatrix}


Similarly


for the matrix B=[111223400]B=\begin{bmatrix} 1 &1&-1 \\ 2 & 2&3\\ 4&0&0 \end{bmatrix}


B1=adjointBdet(B)B^{-1}=\frac{adjointB}{det(B)}


The adjoint of B is calculated by the cofactor method


B1=[0051245840](20)B^{-1}=\frac{\begin{bmatrix} 0 &0&5 \\ 12&4&-5 \\ -8 & 4&0 \end{bmatrix}}{(20)}


B1=[0200205201220420320820420020]B^{-1}=\begin{bmatrix} \frac{0}{20} &\frac{0}{20}&\frac{5}{20} \\ \frac{12}{20} & \frac{4}{20}&\frac{3}{20}\\ \frac{-8}{20}&\frac{4}{20}&\frac{0}{20} \end{bmatrix}


B1=[000.250.60.20.250.40.20]B^{-1}=\begin{bmatrix} 0 &0 & 0.25\\ 0.6 & 0.2 & -0.25\\ -0.4 & 0.2 &0 \end{bmatrix}


Solution (A)



Solution (B)





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