Answer to Question #311850 in Discrete Mathematics for Showi

Question

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+ 3B^T

2.Find B^-1+ ( A^T+A^-1)

Answer 1

Given that

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

Then

$A^{T}=\begin{bmatrix} 1 & 2 & 4\\ 1 & 2 & 0\\ -1 & 3 &0 \end{bmatrix}$ and $B^{T}=\begin{bmatrix} 1 & 2 & 4\\ 1 & 2 & 0\\ -1 & 3 &0 \end{bmatrix}$

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

$det(A)=(1)(0-0)-(1)(0-12)+(-1)(0-8)$

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

$det(A)=20$

And

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

$det(B)=(1)(0-0)-(1)(0-12)+(-1)(0-8)$

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

$det(B)=20$

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

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

The adjoint of A is calculated by the cofactor method

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

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

$A^{-1}=\begin{bmatrix} 0 &0 & 0.25\\ 0.6 & 0.2 & -0.25\\ -0.4 & 0.2 &0 \end{bmatrix}$

Similarly

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

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

The adjoint of B is calculated by the cofactor method

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

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

$B^{-1}=\begin{bmatrix} 0 &0 & 0.25\\ 0.6 & 0.2 & -0.25\\ -0.4 & 0.2 &0 \end{bmatrix}$

Solution (A)

Solution (B)