- #1

thomas49th

- 655

- 0

## Homework Statement

Use the determinate method and also the Guass elimination method to find the inverse of the following matrix. Check your results by direct multiplication

[tex] A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right | = [/tex]

Let's do Guass first

## Homework Equations

Place A by I and attemp to get A into I. Everything I perform on A must be performed on I and when A is in I, the original I is the inverse?

[tex] A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right | [/tex]

[tex] I=\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right | [/tex]

## The Attempt at a Solution

Interchange rows 1 and R

R1 <-> R2

[tex] A =\left | \begin{array}{ccc} 1&0&0\\ 2&1&0\\ 4&1&2 \end{array}\right | = [/tex]

[tex] I =\left | \begin{array}{ccc} 0&1&0\\ 1&0&0\\ 0&0&1 \end{array}\right | [/tex]

Now R2 - 2R1

[tex] A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 4&1&2 \end{array}\right | = [/tex]

[tex] I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ 0&0&1 \end{array}\right | [/tex]

R3 - 4R1

[tex] A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&1&2 \end{array}\right | = [/tex]

[tex] I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ 0&-4&1 \end{array}\right | [/tex]

R3 - R2

[tex] A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&2 \end{array}\right | = [/tex]

[tex] I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ -1&-4&1 \end{array}\right | [/tex]

R3 / 2

[tex] A =\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right | = [/tex]

[tex] I =\left | \begin{array}{ccc} 0&1&0\\ 1&-2&0\\ -1/2&-2&1/2 \end{array}\right | [/tex]

but multiplying my A and new I together does not give me I? What have I done wrong?