Find the determinant using row operations

AI Thread Summary
The discussion focuses on calculating the determinant of a 3x3 matrix using row operations. A participant initially calculated the determinant as -15 but was confused about why it wasn't -30 after multiplying a row by 2. It was clarified that adding a multiple of one row to another does not change the determinant's value, while replacing a row with a multiple of itself does. The correct answer remains -15, and the key takeaway is understanding the different effects of various row operations on the determinant. The conversation emphasizes the importance of distinguishing between these operations in determinant calculations.
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Homework Statement


find the determinant using row operations:
1 -2 2
0 5 -1
2 -4 1



Homework Equations





The Attempt at a Solution


i took row 3 and took 2 x row 1 away from it to get :
1 -2 2
0 5 -1
0 0 -3
1 x 5 x (-3) = -15...but i multiplied a row by 2 so i should get -30 for the det right?but the answer in my book is -15..what am i doing wrong?
 
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Calculating the determinant for A, a 3x3 matrix with elements:
a b c
d e f
g h i

Det(A) = a(ei - fh) - b(di - fg) + c(dh-eg), by starting with row 1.

So -15 is the answer you should be getting.
 


Adding a row or a multiple of a row to another row doesn't change the value of the determinant. If you had replaced a row by a multiple of itself, then the determinant's value would have changed.
 


hmmm well i looked over my book, and it said if i multiply a row by k...i multiply the determinant by k...so when do we multiply the determinant by k?..cause what i did was that not multiplying a row by k(2)?

and thanks for ur help guys
 


vs55 said:
hmmm well i looked over my book, and it said if i multiply a row by k...i multiply the determinant by k...so when do we multiply the determinant by k?..cause what i did was that not multiplying a row by k(2)?

and thanks for ur help guys
You did not replace a row by some multiple of itself; you added a multiple of a row to another row. These are different operations. On the other hand, if you had replaced row 1 by -2 times itself, and then added the first row to the third row, then your determinant would have been +30. This is not what you did though, since the first row stayed the same from start to finish.

In one of the linear algebra books I have, there is a theorem about determinants and row operations. The theorem has three parts.
  1. If you interchange two rows, the determinant of the new matrix will be -1 times the determinant of the old matrix.
  2. If you replace a row by k times itself, the determinant of the new matrix will be k times the determinant of the old matrix.
  3. If you add k times one row to another row, the determinant of the new matrix will be equal to the determinant of the old matrix.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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