Invertible Square Matrices and Determinants - Problem 1
Let's use our result about determinants and the inverse of a matrix to solve another problem. Remember that result says; a square matrix a is invertible if and only if its determinant is not 0.
Let's see if this matrix is invertible, it's a 3 by 3. A way to check is to take its determinant. 3 by 3 determinant is harder than a 2 by 2. Let me write this out 3 4 1, 3 4 1, 4 6 2. Remember when you're calculating determinants you can simplify first using the row or column operations. For example, I could get some 0's by multiplying this column by -1 and adding it to the second column. Let me do that. That's not going to affect the first or last column, so I'll just copy those down 3 4 1, 4 6 2.
Then -3 plus 3 is going to be 0. -4 plus 4 is going to be 0. -1 plus 1 is going to be 0. These two are equal determinants. Let me just expand this determinant along the middle column. This column right here. Don't forget the pattern; plus minus plus, minus plus minus, plus minus plus.
I'm going down the middle column so I'm going to need minus 0 times whatever plus 0 minus 0. Actually I'm seeing here if I expand down this column, I'm going to get a bunch of 0's. 0 times some minor; that's minus 0, plus this 0 times some other minor, minus this 0 times a minor, it's just all going to be 0. What that tells me is that this matrix is not invertible.
Let's see if we can make some kind of generalisation out of this. Why did we get 0's down this column? Because these two columns are exactly the same. Whenever you have two identical columns, you can multiply one by -1 and add it to the other and get 0's. So whenever a matrix has two identical columns it is not invertible. Further more its determinant will always be 0. Don't forget that; determinant 0, matrix is not invertible when 2 rows or columns are identical.