Answer to Question #108256 in Linear Algebra for havefun7741

Question

[img]https://upload.cc/i1/2020/04/06/lcgHuW.jpg[/img]


question is in photo, R=4

Answer 1

$A=\begin{bmatrix} a & 1&0\\ 1&0&b\\ 0 &b&a \end{bmatrix}, a\neq0$

(a) $A^{-1}:\\ C=\begin{bmatrix} a & 1&0|1&0&0 \\ 1&0&b|0&1&0\\ 0&b&a|0&0&1 \end{bmatrix}$

IIr$\leftrightarrow$ Ir

$C=\begin{bmatrix} 1&0&b|0&1&0\\ a & 1&0|1&0&0 \\ 0&b&a|0&0&1 \end{bmatrix}$

IIr+Ir(-a)

$C=\begin{bmatrix} 1&0&b|0&1&0\\ 0 & 1&-ab|1&-a&0 \\ 0&b&a|0&0&1 \end{bmatrix}$

IIIr+IIr(-b)

$C=\begin{bmatrix} 1&0&b|0&1&0\\ 0 & 1&-ab|1&-a&0 \\ 0&0&a+ab^2|-b&ab&1 \end{bmatrix}$

IIIr$\cdot\frac{1}{a(1+b^2)}$

$C=\begin{bmatrix} 1&0&b|0&1&0\\ 0 & 1&-ab|1&-a&0 \\ 0&0&1|\frac{-b}{a(1+b^2)} &\frac{b}{(1+b^2)} &\frac{1}{a(1+b^2)} \end{bmatrix}$

Ir+IIIr(-b)

IIr+IIIr(ab)

$C=\begin{bmatrix} 1&0&0|\frac{b^2}{a(1+b^2)}&\frac{1}{1+b^2} &\frac{-b}{a(1+b^2)}\\ 0 & 1&0|\frac{1}{1+b^2} &\frac{-a}{1+b^2} &\frac{b}{(1+b^2)} \\ 0&0&1|\frac{-b}{a(1+b^2)} &\frac{b}{(1+b^2)}& \frac{1}{a (1+b^2)} \end{bmatrix}$

$A^{-1}=\begin{bmatrix} \frac{b^2}{a(1+b^2)}&\frac{1}{1+b^2} &\frac{-b}{a(1+b^2)}\\ \frac{1}{1+b^2} &\frac{-a}{1+b^2} &\frac{b}{(1+b^2)} \\ \frac{-b}{a(1+b^2)} &\frac{b}{(1+b^2)}& \frac{1}{a (1+b^2)} \end{bmatrix}$

$detA=a(1+b^2)$

(b) R=4, R+2=6

$\begin{Bmatrix} x_1-3x_3=6 \\ 2x_1+x_2=1\\ 9x_2-6x_3=-2 \end{Bmatrix}$

$A=\begin{bmatrix} 1&0&-3 \\ 2&1&0\\ 0&9&-6 \end{bmatrix} B=\begin{bmatrix} 6 \\ 1\\-2 \end{bmatrix} X=\begin{bmatrix} x_1\\ x_2\\x_3\end{bmatrix}\\ AX=B\\ X=A^{-1}B\\ C=\begin{bmatrix} 1 & 0&-3|1&0&0 \\ 2 &1&0|0&1&0\\ 0&9&-6|0&0&1 \end{bmatrix}$

IIr+Ir(-2)

$C=\begin{bmatrix} 1 & 0&-3|1&0&0 \\ 0 &1&6|-2&1&0\\ 0&9&-6|0&0&1 \end{bmatrix}$

IIIr+IIr(-9)

$C=\begin{bmatrix} 1 & 0&-3|1&0&0 \\ 0 &1&6|-2&1&0\\ 0&0&-60|18&-9 &1 \end{bmatrix}$

IIIr$\cdot\frac{1}{-60}$

$C=\begin{bmatrix} 1 & 0&-3|1&0&0 \\ 0 &1&6|-2&1&0\\ 0&0&1|-\frac{3}{10} &\frac{3}{20}&-\frac{1}{60} \end{bmatrix}$

Ir+IIIr(3)

IIr+IIIr(-6)

$C=\begin{bmatrix} 1 & 0&0|\frac{1}{10}&\frac{9}{20}&-\frac{1}{20} \\ 0 &1&0|-\frac{1}{5}&-\frac{1}{10}&-\frac{1}{60} \\ 0&0&1|-\frac{3}{10} &\frac{3}{20}&-\frac{1}{60} \end{bmatrix}$

$A^{-1}=\begin{bmatrix} \frac{1}{10}&\frac{9}{20}&-\frac{1}{20} \\ -\frac{1}{5}&-\frac{1}{10}&-\frac{1}{60} \\ -\frac{3}{10} &\frac{3}{20}&-\frac{1}{60} \end{bmatrix}$

$X=\begin{bmatrix} \frac{1}{10}&\frac{9}{20}&-\frac{1}{20} \\ -\frac{1}{5}&-\frac{1}{10}&-\frac{1}{60} \\ -\frac{3}{10} &\frac{3}{20}&-\frac{1}{60} \end{bmatrix}\cdot \begin{bmatrix} 6 \\1\\-2 \end{bmatrix}=\\ =\begin{bmatrix} \frac{23}{20}\\ -\frac{38}{30}\\ -\frac{97}{60} \end{bmatrix}$

(c)

$det(adj(AB^{-1}))=det(adj(A))det(adj(B^{-1}))\\ det(adj(A))=det(A)=a(1+b^2)\\ det(B)=R+3=7\\ det(adj(B^{-1}))=\frac{1}{7}\\ det(adj(AB^{-1}))=\frac{a(1+b^2)}{7}$

Question 3

$A=[a_1\quad a_2\quad a_3]=\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}& a_{32}&a_{33} \end{bmatrix}\\ detA\neq0\\ B=[2a_1+4a_2-2a_3\quad -a_1-4a_2+3a_3\\ \quad a_2-a_3\quad 3a_1-2a_2+6a_3]=\\ =\begin{bmatrix} b_1\quad b_2 \quad b_3 \quad b_4 \end{bmatrix}$

$b_1=\begin{bmatrix} 2a_{11}+4a_{12}-2a_{13}\\ 2a_{21}+4a_{22}-2a_{23}\\ 2a_{31}+4a_{32}-2a_{33}\\ \end{bmatrix}\\ b_2=\begin{bmatrix} -a_{11}-4a_{12}+3a_{13}\\ -a_{21}-4a_{22}+3a_{23}\\ -a_{31}-4a_{32}+3a_{33} \end{bmatrix}\\ b_3=\begin{bmatrix} a_{12}-a_{13}\\ a_{22}-a_{23}\\ a_{32}-a_{33} \end{bmatrix}\\ b_4=\begin{bmatrix} 3a_{11}-2a_{12}+6a_{13}\\ 3a_{21}-2a_{22}+6a_{23}\\ 3a_{31}-2a_{32}+6a_{33} \end{bmatrix}$

$Bx=b\\ B:3\times4\\ b:3 \times1\\ x:3\times1$