Introduction

I tried to solve the problem C19 - page 205 - A First course in Linear Algebra by Rober A. Beezer.

The question asks to find the inverse of a matrix if exits and verifying it.

I solved the problem using hand and verify the solution using matlib package in r to get the inverse.

We know that AI = inv(A)

1

We need to multiply by the Identy matrix to find the inverse in an augmented matrix

\[\left\{ { \begin{matrix} 1 & 3 & 1 \\ 0 & 2 & 1 \\ 2 & 2 & 1 \end{matrix} }|{ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} } \right\} \quad \]

2

Determine the pivot entry point, which is in this case is 1. We need to eleminate elements under the pivot using the Upper triangle and lower triangle method.

\[{ pivot }_{ 1 }\quad =\quad 1\\ { multiplier }_{ 1 }\quad =\quad \frac { 2 }{ 1 } \\ R3\quad =\quad R3\quad -\quad { multiplier }_{ 1 }\quad \times \quad R1\\ { pivot }_{ 2 }\quad =\quad 2\\ { multiplier }_{ 2 }\quad =\quad \frac { -4 }{ 2 } \\ R3\quad =\quad R3\quad -\quad { multiplier }_{ 1 }\quad \times \quad R1\\ \left\{ { \begin{matrix} 1 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & -4 & -1 \end{matrix} }|{ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1 \end{matrix} } \right\} \quad \Longrightarrow \quad \left\{ { \begin{matrix} 1 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} }|{ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 2 & 1 \end{matrix} } \right\} \]

\[{ pivot }_{ 3 }\quad =\quad 1\]

3

We eliminate the elements above the pivots - Lower triangle.

\[\left\{ { \begin{matrix} 1 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} }|{ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 2 & 1 \end{matrix} } \right\} \quad \Longrightarrow \quad \left\{ { \begin{matrix} 1 & 0 & \frac { -1 }{ 2 } \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} }|{ \begin{matrix} 0 & \frac { -3 }{ 2 } & 0 \\ 0 & 1 & 0 \\ -2 & 2 & 1 \end{matrix} } \right\} \]

\[divide\quad each\quad row\quad by\quad the\quad pivot\quad coeff.\\ \Longrightarrow \quad \left\{ { \begin{matrix} 1 & 0 & \frac { -1 }{ 2 } \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{matrix} }|{ \begin{matrix} 1 & \frac { -3 }{ 2 } & 0 \\ 2 & -1 & -1 \\ -2 & 2 & 1 \end{matrix} } \right\} \quad \Longrightarrow \quad \left\{ { \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} }|{ \begin{matrix} 0 & \frac { -1 }{ 2 } & \frac { 1 }{ 2 } \\ 2 & -\frac { 1 }{ 2 } & -\frac { 1 }{ 2 } \\ -2 & 2 & 1 \end{matrix} } \right\} \]

\[The\quad inverse\quad should\quad be\quad \\ \\ \left[ \begin{matrix} 0 & \frac { -1 }{ 2 } & \frac { 1 }{ 2 } \\ 2 & -\frac { 1 }{ 2 } & -\frac { 1 }{ 2 } \\ -2 & 2 & 1 \end{matrix} \right] \]

Validation in R

To validate the handsolution, I used the matlib package to get the inverse in r

library(matlib)
## Warning in rgl.init(initValue, onlyNULL): RGL: unable to open X11 display
## Warning: 'rgl_init' failed, running with rgl.useNULL = TRUE
a = matrix(c(1,3,1,
             0,2,1,
             2,2,1), nrow = 3, byrow = TRUE)
ai <- inv(a)
ai
##      [,1] [,2] [,3]
## [1,]    0 -0.5  0.5
## [2,]    1 -0.5 -0.5
## [3,]   -2  2.0  1.0