Answer to Question #105340 in Linear Algebra for whitebird

Question

[img]https://upload.cc/i1/2020/03/13/UfFVyi.jpg[/img]



R is 4

Answer 1

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

Solution:

(a)Find the inverse of A:

(I)inverion algorithm

$\begin{bmatrix} 1&4&3|1&0&0 \\ -1&-2&0|0&1&0\\ 2&2&-2|0&0&1 \end{bmatrix}$

$\begin{bmatrix} 1&4& 3|1&0&0\\ 0&2&3|0&1&0\\ 0&-6&-8|-2&0&1 \end{bmatrix}$

Thus

$\begin{bmatrix} 1&4&3|1&0&0 \\ 0&1&1.5|0.5&0.5&0\\ 0&-6&-8|-2&0&1 \end{bmatrix}$

$\begin{bmatrix} 1&0&-3|-1&-2&0 \\ 0&1&1.5|0.5&0.5&0\\ 0&0&1|1&3&1 \end{bmatrix}$

$\begin{bmatrix} 1&0&0|2&7&3 \\ 0&1&0|-1&-4&-1.5\\ 0&0&1|1&3&1 \end{bmatrix}$

Anser:

$A^{-1}=\begin{bmatrix} 2&7&3 \\ -1&-4&-1.5\\ 1&3&1 \end{bmatrix}$

(II)adjoin method

Thus we have $det(A)=2\not=0,$ and hence $A$ is invretible.

To find the inverse using the formula, we first determine the confactors $C_{ij}\space of\space A.$

We have

$C_{11}=4,C_{12}=-2,C_{13}=2,$

$C_{21}=14,C_{22}=-8,C_{23}=6,$

$C_{13}=6,C_{32}=-3,C_{33}=2.$

The adjoin matrix of $A$ is

$Adj(A)=C^T=\begin{bmatrix} 4&14&6 \\ -2&-8&-3\\ 2&6&2 \end{bmatrix}$

$A^{-1}=\frac{1}{det(A)}Adj(A)=\begin{bmatrix} 2&7&3 \\ -1&-4&-1.5\\ 1&3&1 \end{bmatrix}$

(b)Use matrix method to solve each of the folloing

linear systems:

(I)$\begin{cases} x_1+4x_2+3x_3=3, \\ -x_1-x_2+x_3=5,\\ -3x_1-6x_2=2. \end{cases}$

$A=\begin{bmatrix} 1&4&3 \\ -1&-1&1\\ -3&-6&0 \end{bmatrix},d=\begin{bmatrix} 3\\ 5\\ 2 \end{bmatrix}$

$A^{-1}=\begin{bmatrix} 2&-6&\frac{7}{3} \\ -1&3&-\frac{4}{3}\\ 1&-2&1 \end{bmatrix}$

$\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}=A^{-1}*d=\begin{bmatrix} -\frac{58}{3} \\ \frac{28}{3}\\ -5 \end{bmatrix}$

$x_1=-\frac{58}{3},x_2=\frac{28}{3},x_3=-5.$

(II)$\begin{cases} 3x_1+4x_2+x_3=-5, \\ -2x_2-x_3=3,\\ -2x_1+2x_2+2x_3=-1. \end{cases}$

$A=\begin{bmatrix} 3&4&1 \\ 0&-2&-1\\ -2&2&2 \end{bmatrix},d=\begin{bmatrix} -5 \\ 3\\ -1 \end{bmatrix}$

$x=A^{-1}*d;$

$A^{-1}=\begin{bmatrix} 1&3&1 \\ -1&-4&-1.5\\ 2&7&3 \end{bmatrix}$

$\begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}=A^{-1}*d=$

$\begin{bmatrix} 1&3&1\\ -1&-4&-1.5\\ 2&7&3 \end{bmatrix}*\begin{bmatrix} -5 \\ 3\\ -1 \end{bmatrix}=\begin{bmatrix} 3 \\ -5.5\\ 8 \end{bmatrix}$

Answer:

$x_1=3,x_2=-5.5,x_3=8.$

(c)Use Cramer's rule to solve the system of linear

equation in (b)(I)

$A=\begin{bmatrix} 1&4&3\\ -1&-1&1\\ -3&-6&0 \end{bmatrix},d=\begin{bmatrix} 3 \\ 5\\ 2 \end{bmatrix}$

$D=\begin{vmatrix} 1&4&3 \\ -1&-1&1\\ -3&-6&0 \end{vmatrix}=3$

$D_{x_1}=\begin{vmatrix} 3&4&3 \\ 5&-1&1\\ 2&-6&0 \end{vmatrix}=-58$

$D_{x_2}=\begin{vmatrix} 1&3&3 \\ -1&5&1\\ -3&2&0 \end{vmatrix}=28$

$D_{x_3}=\begin{vmatrix} 1&4&3\\ -1&-1&5\\ -6&-6&2 \end{vmatrix}=-15$

$x_1=\frac{D_{x_1}}{D}=\frac{-58}{3},$

$x_2=\frac{D_{x_2}}{D}=\frac{28}{3},$

$x_3=\frac{D_{x_3}}{D}=\frac{-15}{5}=-5.$

(d)Find $det(adj(A))$

$adj(A)=\begin{bmatrix} 4&14&6 \\ -2&-8&-3\\ 2&6&-2 \end{bmatrix}$

$det(adj(A))=4$

(e)Use result of (a) to find $adj((R+2)A^T),R=4$

$A^T=\begin{bmatrix} 1&-1&2 \\ 4&-2&2\\ 3&0&-2 \end{bmatrix}*6=\begin{bmatrix} 6&-6&12 \\ 24&-12&12\\ 18&0&-12 \end{bmatrix}$

$adj((R+2)A^T)=\begin{bmatrix} 144&-72&72\\ 504&-288&216\\ 216&-108&72 \end{bmatrix}$