hello welcome to section 3.2 properties of determinants in the previous section 3.1 we learned how to do determinants by cofactor expansion in this section we're going to add in our row operations to reduce the number of operational cost in cofactor expansion the operational cost is approximately n factorial for products and for a reasonable size matrix say it 50 by 50 matrix this would be approximately 50 factorial or about 10 to the 64 products which is unreasonable and can never be done that way so if you incorporate the properties if you incorporate the row operations into computing your determinants this will significantly reduce the cost down and can quickly have you calculate the determinants so theorem 3 here lists our three property rules notice that the first property rule here is alpha Rho I plus Rho J replace Rho J this does not change to determine at all and this can be used to introduce more zeros into your matrix and significantly reduced your operational cost what I do want to point out to everyone is that do not add in a product here that is do not add in a number that you product this row some students will do that in order to reduce fractions that is they do not encounter a fraction so they'll start adding in numbers there don't do that it will definitely change the determinant so you need to use the rule as we learned it and as it stated here the other rule is if two rows of a are interchanged row I interchange with row J this introduces a negative 1 into the product the last one here is if a row a is multiplied by a constant then that changes the determinant by the constant this one here you have to be very careful with and is illustrated at the bottom of the slide so we'll come back to that one in a moment so the idea is to combine the cofactor that you learned in Section 3.1 with row ops this example here illustrates it for us so if you look at this example we would like to expand here at Row 2 because it has a lot of zeros so the first thing we would do is you would take the cofactor expansion which would start and Row 2 and you would delete this row in this column and you can see you're left with the matrix 1 you left at this matrix here 1 3 4 2 6 10 to 7 11 5 and then it's negative 1 raised to the 2 plus 2 so now you you're working with this matrix to go from this matrix to this one we would do the row op minus 2 Row 1 plus Row 2 replace Row 2 that gives you this new row right here and does not fat excuse me does not change the determinant at all the next thing we would do so we're just gonna copy the matrix down here again the next thing we would do is we to interchange Row 2 with Row 3 that gives us this new matrix which happens to be a upper triangular matrix and the determined it would be minus 5 times the product of the diagonals which is 1 times 1 times 2 for a final answer of minus 10 this significantly reduces the cost particularly that it introduced at this point right here a row that had several zeros into it notice this property which is properties see this is a little this is a little tricky this row op does not follow the normal row op of Rho alpha Rho I replace row I instead you will just factor out the constant and this is illustrated in the next example so let's practice that so here's a 3x3 so this is a simple example which can be done by cofactor expansion but we're going to use row ops and do in our calculation so we wish to compute the determinant of this matrix so the first step here would be is I will pull out the 2 I'm going to factor the 2 out of the first row so you can see it right there the 2 is factored out then they go from this matrix to this matrix will be minus 5 Row 1 plus Row 2 replace Row 2 and you can in it you can see what happens right here is the new row and then I'm also going to do minus 7 Row 1 plus Row 3 replace Row 3 and you can see I also get this new row so I've introduced a column that has a 1 and a bunch of zeros so the next part is I'm going to factor out a minus 4 out of this row and then I'm going to use that row to zero out the minus 8 so they get this new matrix will be eight rode to replace Row three Plus Row 3 replace Row 3 that introduced is the 0 here and you can see I have a 2 times negative 4 times the product of the diagonal elements which gives me to determine a negative 40 so this is a using the combination of cofactor and row operations in doing so I'm my idea is to introduce zeros and using those to quickly calculate so let's do another example so looking at this example I see this column right here has a lot of zeros in it so I'm going to use the cofactor expansion and expand down column 3 so we're starting off with the matrix we would delete column 3 Row 3 and you can see then we end up with the matrix 2 3 1 4 7 3 1 2 4 negative 2 and don't always forget you have this negative 1 raised to the place where the negative 2 is located which happens to be the 3 plus 3 locations so that's a positive number the next thing we're going to do is we're going to introduce a 0 so we're going to do the row op of -2 Row 1 plus Row 2 replace Row 2 in this Center this changes this row right here and introduces a 0 the next thing is we're going to take and do swap Row 2 in doing that swap we introduce a negative 1 so you see that the minus 2 and the 2 have changed so it's now introduced 2 minus 1 and then we're going to roast WAPA ghin to get this new matrix which would be Row one swapped with - that gives us a one at the spot and now I'm going to do the row operation of - - Row 1 plus Row 2 replace Row 2 and that introduces another 0 and I'm going to do one more row op I'm going to do Row 2 plus Row 3 replace Row 3 and it is now in upper triangular form and the determinant will then be negative 2 times the product of diagonal elements for a final answer of negative 1 alternately you could have at this stage did a cofactor expansion down column 1 and there would have been negative 2 times the a11 position 1 times the determinant deleting this row of this column it would have been negative 1 negative 7 1 1 and you would ended up with negative 12 also so you do the row operations to introduce as many zeros as possible but anytime you can stop and do the cofactor expansion if you wish your the idea again is 2 in it is to reduce the amount of calculations so going to the next row 3 age here we're going to for this course we're going to skip this part right here we understand what's going on so we're not going to go through and talk about upper triangular matrices and these determinants in this form what I do want to stress on this slide is this theorem right here this is the the key to what we're looking at is if a matrix is invertible if and only if the determinant matrix is nonzero this is very important it is the main point of both these sections so we're focusing on the determinant and we're trying to determine if you can find an inverse of a matrix so if the determinant is nonzero then you have the inverses that you can find the inverse if the determinant is zero then you cannot find the inverse in one idea to think about here is going back to your 2x2 matrix and notice that if your a matrix is a b c d then a inverse was one over ad minus BC times D minus B minus C a and we were calculating a determinate of a 2x2 to be ad minus BC and if this was zero we can't do this division and we have a problem so the inverse a a inverse does not exist and if this determinant of a was nonzero then the a inverse exists so this is probably the most important part of this section is that if a matrix is invertible if and only if determinant a is not zero the other important theorem here on this line is that the determinant of a transpose will be the same as a determinate of a and this makes sense if you go back to the cofactor if you can expand across any row or are down any column then it makes sense that if you just transpose the matrix then you would end up with the same we're not going to go through the proof of that the final slide for us this is still part of the proof is this property right here that we need to know is that the determinant of a B equals the determinant of a times the determinant of B this is the multiplication property it's very important to told to use one thing I do want to point out is that this does not extend to addition that is the determinate of a plus B will not be equal to the determinant of a plus the determinate of being there are some examples that are listed here to illustrate what's going on so our example the determinate of a cubed given determine a equals five since you can use the product rule then determine if a cubed was determinate of the three a times a times a this will just be 5 times 5 times 5 or 125 another use of this is if you have the inverse of the matrix you can actually calculate the determinant of a inverse which knowing that a inverse times a equals the identity take the determinant of both sides the determinate of identity will be 1 because it's the product of diagonal elements which equals our determinant of a inverse times a using this property this equals the determinant of a inverse times the determinant of a apologize ran out of room and if you divide out this number you will have that the determinant of a inverse equals 1 over the determinant of a so if you know the determinate of a you can find the determinant of a inverse notice that if they determine if a equals 0 then a inverse doesn't exist in this calculation will not be done so this ends our discussion on properties of determinants please practice city of terminus um there is a a practice sheet that you can use these row operations on to do your calculations and that'll be it for this