That is, if you have matrix A:<\/p>\n
![](\"\/..\/images\/deter2_feb16_2018.png\")
<\/p>\n
then<\/p>\n
![](\"\/..\/images\/deter3_feb16_2018.png\")
<\/p>\n
Theoretical point 1<\/strong>:<\/p>\n You can expand the determinant along any row or column. In yesterday’s example, I showed how to evaluate the determinant of B<\/p>\n from the top row. However, you could evaluate det B from the middle column instead. For the negative flip-flopping, remember to multiply each step by (-1)^(r+c), where r is the row, and c is the column.<\/p>\n I’ll now evaluate det B from the middle column (using the procedure from my previous post):<\/p>\n The middle column starts at 11, which is in row 1, column 2. Therefore, the “flip-flop” factor will be (-1)^(1+2)=-1. Imagining the matrix without the first row and second column, we proceed:<\/p>\n -1×11(4*5-(-1)*(-7))=-11(13)=-143<\/p>\n We move to the next number in the second column: the 2. Its flip-flop factor is (-1)^(2+2)=1:<\/p>\n 1×2(3*5 – (-1)(1))=2(16)=32<\/p>\n Now, we arrive at the third member of the middle column, which is a 0. Here we get a break: 0 times anything else is 0.<\/p>\n Finally, we add our results:<\/p>\n det B=-143+32+0=-111<\/p>\n So, det B comes to -111, just as it did from yesterday’s expansion along the top row.<\/p>\n Having the freedom to choose the row or column to expand from is definitely an advantage when evaluating the determinant, since you can make convenient use of zeros in a matrix.<\/p>\n I hope this helps all you college\/university students, for whom first term exams draw near:)<\/p>\n Source<\/em>: Johnson|Riess|Arnold. Introduction to Linear Algebra, 2nd Edition. Don Mills, Ontario: 1989.<\/p>\n![](\"\/..\/images\/deter1_feb16_2018.png\")
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