![](\"\/..\/images\/deter0_feb16_2018.png\")
<\/p>\n
Solution:<\/p>\n
Multiply the numbers along the diagonal from top left to bottom right. Take that result, then subtract from it the product of the other diagonal:<\/p>\n
det A = -2*-9 – 3*4 = 18-12=6<\/p>\n
That’s fine – but what about a larger matrix? In fact, 3×3 matrices are probably the most common ones on which to evaluate the determinant. How do you do it, in that case?<\/p>\n
Example 2: Find the determinant of matrix B: Solution:<\/strong><\/p>\n There are many ways to do this; here might be the most common:<\/p>\n Start at the top left. Imagine the square 2×2 matrix that results by omitting the first row and column (let’s call it matrix P). Multiply the top left number by det P:<\/p>\n 3(2*5 – 0*(-7)) = 30<\/p>\n Continue with the top middle number in B: the 11. Now imagine the 2×2 matrix you get by omitting the top row and middle column. Do it the same as before, except you multiply it by -1 (the process flipflops between 1 and -1):<\/p>\n -1(11)(4*5 – (-1)(-7))= -11(13)=-143<\/p>\n Next step: repeat the process, this time from the top right number. The -1 from last step flip-flops back to 1. We proceed as follows:<\/p>\n 1(4*0 – (-1)(2))=2<\/p>\n Finally, we take our three results from above and add them together:<\/p>\n 30 – 143 + 2 = -111<\/p>\n So, the determinant of matrix B, above, is -111.<\/p>\n There is much to discuss about determinants. I’ll be saying much more about them in future posts:)<\/p>\n Source:<\/p>\n Johnson\/Riess\/Arnold: Introduction to Linear Algebra<\/i>. Don Mills: Addison-Wesley, 1989.<\/p>\n
\n<\/strong><\/p>\n![](\"\/..\/images\/deter1_feb16_2018.png\")
<\/p>\n