Finding inverse matrices using Guass approach

[thomas49th]

Homework Statement

Use the determinate method and also the Gauss elimination method to find the inverse of the following matrix. Check your results by direct multiplication
A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right | =

Let's do Gauss 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?

A =\left | \begin{array}{ccc} 2&1&0\\ 1&0&0\\ 4&1&2 \end{array}\right |
I=\left | \begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right |

The Attempt at a Solution

Interchange rows 1 and R
R1 <-> R2

A =\left | \begin{array}{ccc} 1&amp;0&amp;0\\ 2&amp;1&amp;0\\ 4&amp;1&amp;2 \end{array}\right | =
I =\left | \begin{array}{ccc} 0&amp;1&amp;0\\ 1&amp;0&amp;0\\ 0&amp;0&amp;1 \end{array}\right |

Now R2 - 2R1

A =\left | \begin{array}{ccc} 1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 4&amp;1&amp;2 \end{array}\right | =
I =\left | \begin{array}{ccc} 0&amp;1&amp;0\\ 1&amp;-2&amp;0\\ 0&amp;0&amp;1 \end{array}\right |

R3 - 4R1
A =\left | \begin{array}{ccc} 1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 0&amp;1&amp;2 \end{array}\right | =
I =\left | \begin{array}{ccc} 0&amp;1&amp;0\\ 1&amp;-2&amp;0\\ 0&amp;-4&amp;1 \end{array}\right |
R3 - R2
A =\left | \begin{array}{ccc} 1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 0&amp;0&amp;2 \end{array}\right | =
I =\left | \begin{array}{ccc} 0&amp;1&amp;0\\ 1&amp;-2&amp;0\\ -1&amp;-4&amp;1 \end{array}\right |

R3 / 2
A =\left | \begin{array}{ccc} 1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 0&amp;0&amp;1 \end{array}\right | =
I =\left | \begin{array}{ccc} 0&amp;1&amp;0\\ 1&amp;-2&amp;0\\ -1/2&amp;-2&amp;1/2 \end{array}\right |

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

 

[Dick]

Check the R3-R2 step. (-4)-(-2)=(-2) not (-4).

 

[Mark44]

You made a mistake in the calculation right after what you show as R3 - R2.
The -4 in the last row of I should be -2.

 

[thomas49th]

ahh got it working by starting again and doing another combination... made a silly mistake in this jumble!

Thanks for the pointers guys. My other route still took just as long as this one. Do any of you any tricks to improve Gauss Elimination efficiency or is it just experience.

Thanks
Tom

 

[Mark44]

I don't know how you are doing the calculations, but your work was difficult to check with two separate matrices. The usual practice is to create an augmented matrix, starting with A in the left half and I in the right half. After you get to I in the left half, your inverse of A will be in the right half. Each augmented matrix will look something like this.

\left[ \begin{array}{ccccccc}<br /> 2&amp;1&amp;0&amp;|&amp;1&amp;0&amp;0\\<br /> 1&amp;0&amp;0&amp;|&amp;0&amp;1&amp;0\\ <br /> 4&amp;1&amp;2&amp;|&amp;0&amp;0&amp;1<br /> \end{array}\right ]

An "accounting" tip to make it easier to understand what you did is to use a notation that indicates which row changes when you add a multiple of one row to another. One way to do this is fairly verbose R1 <-- R1 - 3R2. You can abbreviate this to R1 - 3R2 if it's understood that it's always the first row listed that is added to. Obviously if you just switch two rows, it doesn't matter which one you list first, and if you replace a row by a multiple of itself, there's only one row involved, so there shouldn't be any confusion about which rows are involved.

Other than that, you row-reduced your matrices the way I would have, so I don't see anything that you could have done that would have economized your efforts.