# Challenge 80: Can You Evaluate This Determinant?

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- Published on Feb 14, 2019
- Congratulations to Walt S, Minh Cong Nguyen, Theodore Leebrant, Varun Shah, Zemlya Drakona, Fernando Peñaherrera, Alex Burns, Fysty, Serengeti Ghasa, and fabio alan for successfully solving the last week's math challenge question! Walt S was the first person to solve the question.

Welcome, everyone! LetsSolveMathProblems presents one weekly math challenge problem every Wednesday (continental U.S. time). To participate, please comment your proposed answer and explanation below (or discuss potential solutions with other viewers), keeping in mind that your comment should be left unedited to win the recognition prize. Up to the first ten people with correct solutions are recognized in the next challenge video. The solution to a challenge problem is posted as a separate video the following Wednesday.

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usclip.net/p/PLpoKXj-PWCbaDXYHES37_zX4O-kCWxguM

pyLz5 months agoI found it more interesting that the nxn-matrix { (X+n*j + i)^(n-1)} (i = 0,...,n-1, j=0,...,n-1) is the same constant integer C_n independent of X. I checked this for small n=2,...,7, but is this true in general?

Further: The deteminant of {(X + a*i + b*j)^(n-1)} is a monomial in a,b with coefficients given by OEIS sequence A176113.

Finally you can consider for functions f,g the determinant of {(X + f(i) + g(j))^(n-1)}, which is also independent of x.

attyfarbuckle7 months ago^{+1}This "determinate of consecutive squares" thing generalizes quite well:

determinate of consecutive squares = -216

determinate of 4 consecutive integers = -2

determinate of 16 consecutive cubes = 48^4

determinate of 25 consecutive 4th powers = 600^5 = 77, 760, 000, 000, 000

etc

attyfarbuckle7 months ago*determinant

alonamaloh7 months agoThe determinant of a 3x3 matrix has 6 terms, and looking at the denominators only, you can extract a common factor of 1/80^3. What's left is a determinant of consecutive squares with signs in a checkerboard, and after some tedious computation, it turns out to always be 216. Taking the cubic root of 216/80^3, we get 6/80, and simplifying, 3/40.

Eshwar D7 months agothe cube root is 3/40 where a is 3 and b is 40. take 1/80^1/5 from row 1,1/80^4/5 from row 2 and 1/80^7/5 from row 3. then take 1/80^1/5 from column 2 and 1/80^2/5 from column three common outside the determinant. Now column 1 is replaced by column 1 + column 2. column 2 is replaced by column 2 + column 3. then replace column 2 as column 2 - column 1 .simplify using a^2-b^2 formula. repeat the same for rows one and two to get 2 zeroes in the determinant and after this expand column 2 wise.The determinant value is 216/80^3 .Now take cube root of the determinant value to get a/b as 3/40

combined efforts of

Eshwar D

Sameeksha kumar

A.J.Surya

Anjali Anand

Arun chockalingam

Young Sandwich7 months ago27/64000. The numerator cancels out to 216 (I replaced 2023 with x and expanded it) and the denominator cancels out to 80^3, they then simplify to 27/64000. The cube root of that is 3/40.

The one time I can do the problem, I see the video late.

Young Sandwich7 months agoa = 3 and b= 40

Freejo Jose7 months ago3/40

attyfarbuckle7 months agoIf we're allowed to use the Vandermonde determinate equation it is a little easier.

1) Multiply the columns by -80^(0.2), 80^(0.4), -80^(0.6) respectively.

2) Multiply R2 by -80^(0.6), & R3 by 80^(1.2)

3) You now have | 2019^2 2020^2 2021^2 |

| 2022^2 ......

Put a = 2019, b = 2022, c = 2025 and do some column operations to get

| a^2 2a + 1 4a + 4 |

| b^2 2b + 1 4b + 4 |

| c^2 2c + 1 4c + 4 |

which equals

| a^2 2a + 1 2 |

| b^2 2b + 1 2 |

| c^2 2c + 1 2 |

which equals

| a^2 2a 2 |

| b^2 2b 2 |

| c^2 2c 2 |

which equals 4 times

| a^2 a 1 |

| b^2 b 1 |

| c^2 c 1 |

4) This final determinate is (a-b)(a-c)(b-c) by the formula, which is -3.-6.-3 = -54

5) The determinate in the question is -54 . 4 / -(80^(0.2 + 0.4 + 0.6 + 0.6 + 1.2)) = 216/80^3

6) Final answer 3/40

CoolBlue Bryan7 months agoa=3, b=40

3/40

Piero Puelles7 months agoI get 3/40 . First let x=2023 and y=80^(1/5) therefore expand the determinate and get f(x)/y^(15) and then evaluate f(0)=216 and finally the cube root us 216^(1/3)/80^(15/15) so get 6/80 =3/40 a and b

Piero Puelles7 months agoIn addition f(x) is a constant function so I choose f(0) to be more easy

Ujjwal Sinha7 months agoCan you solve it??!!.. Oh no.. sry,, u hv already solved it I think.. but no matters keep it on..,,👍👍

Pete Berg7 months ago^{+1}Insert Gauss-Jordan here*

Jeet Sharma7 months agoFirst of all let's reduce the writing cumbersomeness of the Determinant by taking fifth root of 80 as 'K' and 2019 as 'x'. The det reduces to a very simple looking with all the terms quadratic in x and K. Taking 1/k out from first row, 1/k^4 from 2nd row and 1/k^7 from 3rd row. Again taking 1/k out from 2nd column and 1/k^2 out from 3 column, the det resolves to 1/k^15 times det containing quadratic terms. Now, then following operations I performed in order

R3- R3 + R2

R2- R2 + R1

I took 3 common out of both R3 and R2, and then applied C3 - C3 + C2

C2 - C2 + C1 and finally C3 - C3 + C2 and expanding the det from 3rd column I got 9*24/k^15 . Putting the value of K that I let and taking the cube root I got 3/40 as my ans.🙌

ba ke7 months ago27/64000

Typo7 months ago^{+1}@Zυвi oh right, sorry. I stated the wrong part of the question

ba ke7 months agoOh, "cube root" !

The answer is 3/40.

Zυвi7 months agoIt is, 27=3^3 while 3 does not divide 64000

Typo7 months agogcd(27,64000)=/=1

Johannes H7 months agoYou can see that each summand contains 1/ the 5th root of 80^15 therefore you can facor out 1/80 from the third root of the determinant. I then tried to write all of these in terms of 2024 and 2027 using binomials, but that turned out pretty complicated, but I guess, if I had taken only one value (like 2023) and named it x, the resulting expression would have been easier to cancel)... In the end my calculator was far more convinient and the result is 6/80 = 3/40

Георгий7 months ago0 obviously

Muhammad Mazhari7 months ago3/40

UbuntuLinux7 months agoI was too late last time :((

adandap7 months agoI just learned something about the comments in USclip! Text between two minus signs seems to be rendered as crossed out. So ignore that - and I meant to say TWO terms each with x^2, (x-1)^2 and (x+1)^2.

adandap7 months agoThere's probably a faster way, but it's not hard to simply compute the determinant by expanding along the top row. You get a constant factor of (80)^(-3) and a bunch of squares -2019^2 2023^2 2027^2 etc. I set 2023 = x, so 2022 = (x-1) etc. Then e six terms have three with an x^2 term, three with each of (X+- 1). By adding those in pairs and then taking the total sum, lots of nice cancellations occur (I suspect that means I've missed a nice symmetry here!) and you get 9*24. So the determinant = 27/64000, and its cube root is 3/40.

Sourin Chatterjee7 months agoTaking out common row wise (80)^1/5, (80)^4/5,(80)^7/5 from 1st 2nd and 3rd. Then row wise common (80)^1/5and(80)^2/5from 2nd and 3rd. Then left with squares. Keeping the first collum first done C3>C3+C2 and then C2>C2+C1 and get reduced form. 2nd collum-(4039,-4045,4051) 3rd collum-(-4041,4047,-4053). Then again C2>C2+C3 and take out 2 common from 2nd collum. Now R1>R1+R2 and R3>R3+R2. And now get 010 in 2nd collum now calculating value we get 3/40.

mstmar7 months agoanswer: 3/40

first thing i did was factor out all of the fifth roots from each row and column, 80^(1/5) from r1, 80^(4/5) from r2 and 80^(7/5) from r3, leaving only 80^(1/5)s in the 2nd col and 80^(2/5) from the 3rd, which can be taken out. This gives us a total of 80^(15/5) = 80^3 taken out of the denominators which we'll keep aside until the end.

next we can add c2 to c1 and c3, not changing the determinants. each of these additions are easy to simplify as they are differences of squares which simplify nicely eg -2019^2+2020^2 = (2020-2019) * (2019+2020) = 4039 :

[4039, 2020^2, -4041]

[-4045, -2023^2, 4047]

[4051, 2026^2, -4043]

we can do the same thing with rows to get rid of even more squares, again simple simplifications.

[-6 ,-3*(4043),6],

[-4045,-2023^2 ,4047]

[6 ,3*(4049),-6]

Lucky, we can get some 0s by adding r3 to r1 leaving r1 as [0,18,0] and we're basically done.

We could just plug this into the determinant formula right now and not have too much difficulty, but we can get another useful 0 by adding c1 to c3, then swapping rows twice (thus not changing det) to get it triangular:

[6 , 0, 0]

[ _ , 18 , 0]

[_ , _ , 2]

giving us the determinant 6*18*2 = 6^3

putting that together with the 1/80^3 we took out at the beginning, we get the determinant being (6^3/80^3) = (3/40)^3 and its cube root being 3/40

Varun Shah7 months agoTake the 1 / 80^(1/5) terms out.

so taking the 1 / 80^3 term out and C1 => C1 + C2 and C2 => C2 + C3 gives me (4039, - 4045, 4051) as first column and (-4041, 4047, - 4053) as second column. The third column has the denominators as in the question.

Now C1 => C1 + C2 turns the first column into (-2,2,-2). Take the 2 out of the determinant.

Do R1 => R1 + R2; R2 => R2 + R3. This gives (a-3)^2 - a^2 = (2a - 3) * 3. and now the first column is (0, 0, - 1) , the second column is (6, - 6, - 4053), the third column is (3*4045, - 3*4051, - 2027^2)

now do R1 => R1 + R2 again so that the the 6s cancel out and the third column also simplifies.

now upon simplyfying and taking the cube root gives 3/40

Serengeti Ghasa7 months agoWith love and regards

Serengeti Ghasa7 months agoThe answer is 3/40

I found as follows

Multiply power 3/5 of 80 in R1 and devide the same in R3 again multiply 1/5 power in C1 and decided the same in C3 then

1/80 would be common in all row then 1/80³ would come out side of determinant

again

C1+C2=C1 and C2+C3=C2

again R1+R2=R1 and R2+R3=R2

Again R1+R2=R1

I found 1/80³(6×3×4051-6×3×4045)×2

=6³/80³

so the cube root is 6/80 = 3/40

Akhil kushe7 months agoBy using operation and simplifying the determinant ...you finally get 6³/80³....therefore it's cube root will be 6/80 i.e. 3/40 ...therefore a=3 and b=40

LetsSolveMathProblems7 months agoAs of now, your solution is not comprehensive enough to be considered for recognition. Could you specifically mention some of the operations or simplifications you performed? (An example would be "I replaced Row 1 with the sum of Row 1 and Row 2.")

* Also, please do not edit your comment, but post a new comment with the modified solution

張惟淳7 months ago3/40. My process: imgur.com/a/QijYnvr

Jaleb7 months ago^{+4}Let x=2023, and y=80^(1/5). This'll make the matrix into the following:

|(x-4)^2/y^1 (x-3)^2/y^2 (x-2)^2/y^3|

|(x-1)^2/y^4 (x-0)^2/y^5 (x+1)^2/y^5|

|(x+2)^2/y^7 (x+3)^2/y^8 (x+4)^2/y^9|

Taking the derivative cancels all x when expanded and leaves 216/y^15.

Taking the cube root gives 6/y^5 or 6/80 = 3/40

Generally the value of x does not matter.

Ankit Sahani7 months ago@Jaleb can we meet on facebook or instagram??please

Jaleb7 months ago@Ankit Sahani United States

Ankit Sahani7 months ago@Jaleb from which country u hail ??

Ankit Sahani7 months agoNice solution

Ankit Sahani7 months ago@Jaleb ohh..

Andre Ben7 months ago^{+6}Ok so I first factor 1/80^(1/5) on the first row, then 1/80^(2/5) in the 2nd and 1/80^(3/5) in the 3rd. Then I see I can factor 1/80^(3/5) in the 2nd column and 1/80^(6/5) in the 3rd column. In total I get 1/80^((1+2+3+3+6)/5)=1/80^3 and I only need to find the determinant of the original matrix, but wihout the different fifth roots. To do that I add the 2nd column to the first and 3rd. These columns change into (4039, -4045, 4051) (Wish I could do latex here!!) and the 3rd becomes (-4041, 4047, -4053).

Now things are going to simplify very quickly: I take the 3rd column, I add it to the first. The 1st column becomes (-2, 2, -2). Now I take the 2nd line and add it to the 1st and 3rd line, so that the first line becomes (0, -3*4043, 6) and the 3rd (0, 3*4043, -6). I factor a 2 from the 1st column then -3 from the 1st line and another 3 from the 3rd line.

I keep in mind the resulting -18 as a factor.

Now the matrix looks like this:

(0 4043 -2)

(1 -2023^2 4047)

(0 4049 -2)

I use the 1st colmun to eliminate the 4047 and -2023^2; now I substract the 1st line to the 3rd and I get the matrix:

(0 4043 -2)

(1 0 0)

(0 6 0)

I can factor a -2 from the 3rd column, then use this 3rd column to eliminate the remaining 4043. So that I get 36=6^2 times the determinant of

(0 0 1)

(1 0 0)

(0 6 0)

which is 6. So the determinant without the fifth roots is simply 6^3. Thus the whole determinant is 6/80 which is 3/40 in reduced form.

(I edited the mistakes! Sorry for the mess!)

Arun Bharadwaj7 months agoUpon expanding the determinant along row 1, it is very easy to see that the denominator is 80^3. (consider just the exponents of 80. 1st term: 1+(14) = 15. 2nd term: 2+(13) = 15. 3rd term 3+(12) = 15. Taking the 5th root of 80^15 gives 80^3. Now, for the numerators, in term 1, we have the following symmetry (year1)^2(year2+1)^2- (year1+1)^2(year2)^2. In term 2, we have (year1)^2(year2 - 3)^2 - (year1 + 3)^2(year2)^2. And finally in term 3: we have (year1)^2(year2 -1)^2 - (year2)(year1+1)^2. It is easy to apply the difference of squares formula to this, and after exploiting the symmetry, we have various things cancel out. In the end, we have a simple difference of products to compute. After some tedious computation, the numerator simplifies to 216. Therefore, the required solution is cube root of 216/80^3 = 3/40. Therefore, a=3, b=40.

Varun Shah7 months agoThe answer is 3 / 40.

first factor out 80^1 / 5 and its multiples to get 80^3 outside. Now inside the terms are in the form a^2 - b^2 and factor them out to make it a little easier. Then simplify it to get 216.

cube root is 6 / 80 or 3/40.

Varun Shah7 months ago+LetsSolveMathProblems

Well as I've seen other replies,I realise everyone has more or less the same solution; just some steps here and there are different. If you don't mind me asking, is your solution following the same general idea or is it something different? 🙄😀

LetsSolveMathProblems7 months agoYour revised solution is excellent. =)

Varun Shah7 months agoso taking the 1 / 80^3 term out and C1 => C1 + C2 and C2 => C2 + C3 gives me (4039, - 4045, 4051) as first column and (-4041, 4047, - 4053) as second column. The third column has the denominators as in the question.

Now C1 => C1 + C2 turns the first column into (-2,2,-2). Take the 2 out of the determinant.

Do R1 => R1 + R2; R2 => R2 + R3. This gives (a-3)^2 - a^2 = (2a - 3) * 3. and now the first column is (0, 0, - 1) , the second column is (6, - 6, - 4053), the third column is (3*4045, - 3*4051, - 2027^2)

now do R1 => R1 + R2 again so that the the 6s cancel out and the third column also simplifies.

now upon simplyfying and taking the cube root gives 3/40

LetsSolveMathProblems7 months agoAs of now, your solution is not comprehensive enough to be considered for recognition. Could you specifically mention some of the numbers or expressions you have obtained during the calculation? Furthermore, if you used any simplification techniques (other than the difference of squares), please mention those, as well.

* Also, please do not edit your comment, but post a new comment with the modified solution.

Andre Ben7 months agoI found 3/40.

Lots of manipulations, but I start by factoring powers of 1/80^(1/5) along lines then columns then I use the famous a^2-b^2=(a-b)(a+b) multiple times to simplify the remaining determinant, by manipulating the lines and columns

LetsSolveMathProblems7 months agoYou are correct. By convention (if there are no parentheses), 2^2 is evaluated first, then the "-" sign is applied. Thus, -2^2 = -4. If you wish to express the product of -2 with itself, you should write (-2)^2.

Andre Ben7 months ago@Marco Dalla Gasperina As far as I'm concerned yes, powers are prioritized, but I agree that it kinda looks weird. I adopted this notation to avoid putting more paranthesis as I already wrote quite a few

Marco Dalla Gasperina7 months ago^{+1}Tangential question: does a unary minus on a number have lower priority than the exponent? It's obvious by the question that -2^2 means -(2^2). But it just looks weird to me.

LetsSolveMathProblems7 months agoThat should be perfect. Thank you! =)

Andre Ben7 months ago^{+1}@LetsSolveMathProblems Ok ok I'll copy paste it. I hope it'll be ok ^^;

aby p7 months agoThe problem looked quite horrifying but ended up being simple so here goes my approach:

First I subracted the first two columns then used the rule a^2-b^2=(a+b)(a-b) then we again subtract the third and the second column and apply the same rule again then we subtract the first two rows and write (n+3)^2-n^2 therefore and then we again subtract the first two columns and then solve the determinant we get the ans 3/40 a=3 and b=40

Benjamin Wang7 months agoShould be 3/40. First observe you’ll get a 1/80 factor. Then you’re left with the squares, which, with some differences of squares formulas and letting x=2023 you’ll get 3(x-6)^2(2x^2-5)-6(x-5)^2(2x^2-2x-6)+3(x-4)^2(2x^2-4x-3), which turns out to have all coefficients = 0 except the constant term, which is -540+900-144=216. After taking the cube root and simplifying, the result follows.

Avi Uday7 months agoThe answer is 3/40. First, i divided the first column by 80^1/5, and multiplied the determinant by the same and performed a C1 >> C1+C2 operation. Then i divided the second column by 80^1/5 and multiplied the determinant by another 80^1/5 and performed a C2 >>> C2 + C3 operation. Then i divided the first column by 80^1/5 and multiplied the determinant by the same. 80^(-3/5) can be taken common from the first row, 80^(-6/5) from the second row, and 80(-9/5) from the third row. Then performed the R1 >> R1+R2 operation twice and expanded the determinant to get the answer.

Smokie Bear 🔴🔵7 months ago^{+18}I plugged the matrix into an online calculator and I got 27/64000. So the cube root is 3/40 quick maths easy peasy. a=3, b=40

KEn Zhu7 months agoit is blatant for such action will not be accepted

Smokie Bear 🔴🔵7 months ago^{+10}It didn’t rely *only* on electronic devices, I calculated the cube root of 27/64000 in my head lol

LetsSolveMathProblems7 months ago^{+2}A solution that only relies on electronic devices will not be accepted. The only exceptions are the enumerative combinatorics problems, for which I usually accept a solution using a computer program (created by the viewer himself or herself) as long as the key algorithms are clearly explained.

Benjamin Wang7 months agoThanks, your answer turns out to be helpful for checking my answer. Not sure about how messy this whole thing is though. I might be missing a trick

Asad Yamin7 months ago@Smokie Bear 🔴🔵 I arrived at the same conclusion