Compute the inverse of the matrix
- Thread starter frctl
- Start date
D
Deleted member 4993
Guest
Almost. Your technique is correct, but you made two errors. The first is one of intent: What you wrote is 1/ad-bc, which when "translated" literally means \(\displaystyle \frac{d}{a} - bc\), whereas what you meant was \(\displaystyle \frac{1}{ad-bc}\), which can be written in plaintext as 1/(ad - bc). Please be sure you understand why these grouping symbols are not optional.
Secondly, you made a reading error it seems. When plugging the values into \(\displaystyle \frac{1}{ad-bc}\), you somehow concluded that the number corresponding to \(c\) is 2, when it's rather clearly 4 in your matrix. If you clear that up, you'll get a denominator of 20, which will give you the right answer.
Secondly, you made a reading error it seems. When plugging the values into \(\displaystyle \frac{1}{ad-bc}\), you somehow concluded that the number corresponding to \(c\) is 2, when it's rather clearly 4 in your matrix. If you clear that up, you'll get a denominator of 20, which will give you the right answer.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
You could have noticed it and fixed it, before posting. People have asked you a number of times to double-check your work. (You still copy a lot of numbers wrong, and you still make a lot of simple arithmetic mistakes.) I'm curious to know, are you going to start double-checking your work?
It's best to verify that you've copied an exercise, expression, or numbers correctly, before you start calculations.
It's best to double-check arithmetic after each step, not at the end of the exercise.
? These suggestions will save you a lot of time.
As long as I'm here, I'd like to ask you to read this post again. You're still starting threads with no effort shown, and you're asking tutors to do your work too much. Please follow the guidelines. Thank you.
\(\;\)