Matrix Calculations in Linear Algebra

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For example, if we have a system of equations:
x + y = 5
2x y = 8
We can write this as an augmented matrix:
| x | y |
|—-|—-|
| 1 | 1 |
| 2 | -1 |
| 5 | 8 |

Now, to solve for the values of x and y, we use a technique called row operations. Basically, we manipulate the rows of our matrix until it’s in a special form (called reduced row echelon form) that makes solving for the variables easy peasy. Here’s how:

1. Divide the first row by 1 to get rid of the x term:
| 1 | 1 |
|—-|—-|
| 1 | 0 |
| 2 | -1 |
| 5 | 8 |

2. Subtract twice the first row from the second row to get rid of the x term:
| 1 | 1 |
|—-|—-|
| 1 | 0 |
| 0 | -3 |
| 5 | 8 |

3. Divide the second row by -3 to get rid of the y term:
| 1 | 1 |
|—-|—-|
| 1 | 0 |
| 0 | 1 |
| 5 | 8 |

4. Subtract 5 times the first row from the third row to get rid of the x term:
| 1 | 1 |
|—-|—-|
| 1 | 0 |
| 0 | 1 |
| 0 | -3 |

Now, we can easily solve for our variables by looking at the last row. The value of x is whatever number you want to make y equal to (in this case, it’s not really important because there are infinitely many solutions). And the value of y is just -3.

So if we wanted to find a solution where x = 2 and y = 0, that would be our answer! But what about when you have more than two variables? Well, it’s basically the same process but with more rows and columns in your matrix. And instead of solving for just one variable (like x or y), you solve for all of them at once using a technique called Gaussian elimination.

But that’s enough math talk for now! Let’s wrap this up before we get too bored.

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