Challenge: Trace of Inverse and Inverse Squared

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• Published on Aug 23, 2018
• Congratulations to Kevin M, Benjamin Wang, Zain Majumder, attyfarbuckle, Throxs, Beshoy Nabil, Richard Chen, NoName, Allan Lago, and dominofan238 for successfully solving the last week's math challenge question! Kevin M was the first person to solve the question.
Your support is truly a huge encouragement.
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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

• Ankit Sahani 9 months ago

1/det(A) comes out to be the final answer.... Which is a constant ... put detA=2018

• Anthony Gibbons 9 months ago

Give A entries a¹¹, a¹² in the first row and a²¹, a²² in the second row
[a¹¹ a¹², a²¹ a²²]
Determinant of A det(A)=a¹¹a²²-a¹²a²¹=2018
The inverse of A, A^-1 = 1/det(A) [a²² -a¹², -a¹² a¹¹]
Firstly, the trace of this is just the sum of the diagonal entries
tr(A^-1)=1/2018[a²²+a¹¹]
(tr(A^-1))^2 = 1/(2018)^2[(a²²)^2+2a¹¹a²²+(a¹¹)^2]
A^-2=(A^-1)^2 = 1/(2018)^2[(a²²)^2+a¹²a²¹ -a¹²a²²-a¹²a¹¹, -a²¹a²²-a²¹a¹¹ a¹²a²¹+(a¹¹)^2]
The trace is again the sum of the diagonal entries
I.e. tr(A^-2)=1/(2018)^2[(a²²)^2+2a¹²a²¹+(a¹¹)^2]
Taking the difference between these 2, and factoring out the 1/(2018)^2
tr(A^-1))^2-tr(A^-2)= 1/(2018)^2[(a²²)^2+2a¹¹a²²+(a¹¹)^2-(a²²)^2-2a¹²a²¹-(a¹¹)^2]
=1/(2018)^2[2a²²a¹¹-2a¹²a²¹]=1/(2018)^2[2(2018)]=1/1009

• Atharva #breakthrough 9 months ago +1

1/1009. eazy, your 1st prob i ever solved. because after a bunch of algebra, we see that the expression simplifies to 2/ det a = 2/2018 = 1009

• Daniel Darroch 9 months ago +1

The challenge is essentially to show that tr(A^(-1))^2-tr(A^(-2)) = 2/det(A) for a 2x2 invertible matrix A, but it turns out that this can be vastly generalised!
Firstly let's simplify by moving away from inverses by letting M = A^(-1), in which case det(M) = 1/det(A), so the relation becomes tr(M)^2-tr(M^2) = 2det(M), for any 2x2 matrix M.
A relation between products of traces of powers of M and the determinant exists for nxn matrices for all n. In particular:
det(M) = Sum[(-1)^(n+Sum[j_k,{k,1,n}]) Product[tr(M^k)^(j_k),{k,1,n}]/Product[k^(j_k) (j_k)!,{k,1,n}], {j in N^n : Sum[k j_k,{i,1,n}] = n}]
This is a bit tricky to read but basically the sum is over all n-tuples j (so j_k is the kth element of j), where the elements of j are in N (the natural numbers, or the non-negative integers) and such that the sum of the elements of j, weighted by their index, gives n. Then each element of the sum has a tuple j and is given by a sign, a product of powers of traces of powers of M, divided by a combinatorial factor, where each part is dictated by j (or rather its elements).
Note that multiplying through by n! gives an integer relation between the determinant and the products of traces (since the constraint on the tuples forces the combinatorial factor to divide n!).
So our next example of this relation, the 3x3 case, is given by tr(M)^3-3tr(M)tr(M^2)+2tr(M^3) = 6det(M).
Then the 4x4 case is tr(M)^4-6tr(M)^2tr(M^2)+3tr(M^2)^2+8tr(M)tr(M^3)-6tr(M^4) = 24det(M).
As a final note, the coefficients and the combinatorics exactly follows the cycle index of the symmetric groups, but with an added sign dictated by the parity of the number of cycles. The sign of tr(M)^n in the relation is always positive, hence the (-1)^n factor, so the sign of a factor is positive when the parity of the number of cycles is equal to the parity of n.

• renpnal 9 months ago +1

Denote A^-1
=
a b
c d
Now (tr(A^-1)^2 - tr(A^-2))
= (a + d)^2 - ((a^2 + bc) + (bc + d^2))
= 2det(A^-1)
= 2/det(A)
= 2/2018
= 1/1009

• Jack Jones 9 months ago

I think u should increase the level of your questions!!!!!

• Amin Thainat 9 months ago

Yes!! right

• Johannes H 9 months ago

The constant is 1/1009

• D.E. Xyk 9 months ago

It is so easy

• Andrey Kolesnik 9 months ago

1/1009.
Let A = {{a,b},{c,d}}
tr A^(-1) = 1/2018*(a+d)
(tr A^(-1))^2 = 1/(2018)^2*(a+d)^2
tr A^(-2) = 1/(2018)^2*(d^2+2bc+a^2)
So, diff. is 1/(2018)^2*2(ad-bc), but det = 2018, so answer is 2/2018 = 1/1009.

• Joel Courtheyn 9 months ago

The result is 1/1009.
I made a simulation in R programming below, to visualize what happens.
#20180823 Weekly Math Challenge 55 - LetsSolveMathProblems
library("pracma", lib.loc="D:/R/win-library")
A

• First I calculated A(-1) and A(-2) (given the detA=2018) to be
A(-1)={{d, -b}, {-c, a}}*2018^(-1)
And
A(-2)={{d^2+bc, -ab-db}, {-ac-dc, a^2+bc}}*2018^(-2)
Now I started calculating the traces to be
Tr(A(-1))=(a+d)/2018
And
Tr(A(-2))=(d^2+2bc+a^2)/2018^2
Now it’s given that
detA=2018
Substituting that in you get
=((a+d)^2+4036)/2018^2
Now calculating
Tr(A(-1))^2-Tr(A(-2))
=(a+d)^2/2018^2-((a+d)^2+4036)/2018^2
=1/2018^2((a+d)^2-(a+d)^2+4036)
=1/2018^2(4036)
=1/1009~0.00099108

• Thomas Dubroux 9 months ago +2

The answer is 1/1009. You just need to let A=[a b c d], evaluate the differents terms in function of a, b, c and d and make the calculus.
I found this problem very easy compared to the other ones that we have usually.

• Will Bishop 9 months ago +1

det A = 2018
det A^-1 = 1/2018
Let A^-1 = [[a, d], [b, c]]; then A^-2 = [[a^2+bc, ab+bd], [ca+cd, bc+d^2]]
tr(A^-1) = a+d, tr(A^-1)^2 = a^2 + 2ad + d^2, tr(A^-2) = a^2 + 2bc + d^2
tr(A^-1) = 2ad - 2bc = 2 det A^-1 = 1/1009

• Univ3rse.Jungle 9 months ago

En utilisant la comatrice et du calcule on trouve 2/det(A) donc 1/1009
Sorry, I m french and I don't know very well the vocabulary in math :)

• Hyperbolium 9 months ago

Finally; a problem I can tackle!

• Kyle Rachman 9 months ago

Let A = [[a, b], [c, d]]. Det(A) = ad-bc = 2018. By the formula for the inverse of a 2x2 matrix, A^-1 = [[d/(ad-bc), -b/(ad-bc)],[-c/(ad-bc), a/(ad-bc)]] = [[d/2018, -b/2018],[-c/2018, a/2018]]. Then tr(A^-1) = (d+a)/2018. A^-2 = [[(d^2+bc)/2018^2, (-db-ba)/2018^2], [(-cd-ac)/2018^2, (cb+a^2)/2018^2]], so tr(A^-2) = (d^2+2bc+a^2)/2018^2, so (tr(A^-1))^2 - tr(A^-2) = (2ad-2bc)/2018^2 = (ad-bc)/(1009*2018) = 1/1009.

• It's an easy problem.Answer=2/det.A=2/2018=1/1009. Thanks all the Mathematically Inclined persons , and thanks specially to Mr Kenny Yip for method using eigenvalues.

• legonick05 9 months ago +5

I thought about this using a cool theorem called the Cayley-Hamilton theorem, which states that if you substitute a square matrix into its characteristic polynomial, you get the zero matrix.
The characteristic polynomial of any 2x2 matrix M is x^2 - tr(M)x + det(M). Using the Cayley-Hamilton theorem on A^-1 means A^-2 - tr(A^-1)A^-1 + det(A^-1)I is the zero matrix, where I is the 2x2 identity matrix.
Since det(A^-1) = 1/det(A) = 1/2018, some algebra shows (1/2018)I = tr(A^-1)A^-1 - A^-2.
Now we will take the trace of both sides. Since tr(I) = 2 we get tr((1/2018)I) = (1/2018)tr(I) = 2/2018 = 1/1009. In addition, tr(tr(A^-1)A^-1 - A^-2) = (tr(A^-1))^2 - tr(A^-2) from properties of the trace operation.
Thus, (tr(A^-1))^2 - tr(A^-2) = 1/1009.

• LetsSolveMathProblems  9 months ago +2

I actually discovered the identity by using Caley-Hamilton Theorem. I'm glad you pointed it out, although it may not be as straightforward as the approach using eigenvalues. =)

• madhav 's talks 9 months ago

Ans is 1/2009
Here's a solution:)
let x, y be the values of A,
Then:- det(A) =2018
det(A inv) =1/2018
Tr(A inv)²= (1/x+1/y)². (1)
Tr(A inv²) = ( 1/x²+1/y²). (2)
Subtract (2) from (1):-
We get 2/xy= 1/1009:)))

• There are 2 methods
Method1 : Let u,v be the eigenvalue of Matrix A such that u*v=det(A)=2018, tr(A)= u+v
Then we have tr(A^-1)= 1/u+1/v
tr(A^-2)= 1/u^2 + 1/v^2
So
tr(A^-1)^2 - tr(A^-2) = (1/u+1/v)^2 - (1/u^2 + 1/v^2) = 1/u^2 + 2/(u*v)+ 1/v^1 - (1/u^2 + 1/v^2)
=2/(u*v) = 2/det(A) = 1/1019
Method2: Let A=Matrix[(a,b),(c,d)], where a,b,c,d are real numbers
A^-1= 1/2018 * Matrix[(d,-b),(-c,a)]
A^-2= 1/2018^2 * Matrix[(d^2+bc,-bd-ba),(-cd-ac,bc+a^2)
So

• Benjamin Brat 9 months ago +1

The constant value is 1/(1009). I obtained it by computing it for an arbitrary matrix [a b; c d].
The trace of A^(-2) has in its numerator a^2+2*bc+d^2 which can be transformed using the fact that det(A) is 2018 into (a+d)^2-2*2018. When substracting the squared trace of inv(A) nly the constant part remains. The denominator is the same in both cases because it comes from 1/det(A) in the inversion process which is a constant.

• madhav 's talks 9 months ago

@Benjamin Brat yeah, man I did it by that method only:)

• Benjamin Brat 9 months ago

I must say that Kenny Yip's eigenvalue method is prettier and faster though.

• DragonFire17 9 months ago +1

Let A=[a b;c d]
A^-1 = (1/det(A))[d -b; -c a]
A^-2 = (1/(det(A)^2))[d^2-bc -(db+ba);-(cd+ac) bc+a^2]
tr(A^-1)=(d+b)/det(A)
tr(A^-2)=(d^2+2bc+a^2)/(det(A)^2)
Therefore:
= 2det(A)/(det(A)^2)
=2/det(A)
=2/2018
=1/1009

• Erin Cobb 9 months ago

1/1009
Let A=[[a b] [c d]] then |A| = ad-bc = 2018
A^-1=[[d/2018 -b/2018] [-c/2018 a/2018]]
tr(A^-1)=d/2018+a/2018=(d+a)/2018
(A^-2)=(A^-1)^2=[[(d*d-b*-c)/2018^2 (d*-b-b*a)/2018^2] [(-c*d+a*-c)/2018^2 (-c*-b+a*a)/2018^2]]
tr(A^-2) = (d^2+2bc+a^2)/2018^2

• Jonathan Fraser 9 months ago +1

2018 * l1*l2 = 1 from det A = 2018, where l1 and l2 are inverse eigen values (or eigen values of inverse). (tr(A-1))^2 - tr(A-2) = (l1 + l2)^2 - (l1^2 + l2^2) = 2l1l2 = 2/2018 = 1/1009

• Falah Edu 43 9 months ago

Let
A = [a b]
[c d]
and detA = ad-bc = 2018
A^(-1) = [d/detA -b/detA]
[-c/detA a/detA]
A^(-2) = [d^2-bc/(detA)^2 db+ab/(detA)^2]
[-cd-ac/(detA)^2 -cb+a^2/(detA)^2]
so we get :
Tr(A^-2) is d^2-2bc+a^2/(detA)^2
so we get Tr(A^-1)^2 - tr(A^-2) = 2(ad-bc)/(detA)^2 = 2/detA
it proves that for any a,b,c,d the final answer will constant, because the detA is constant detA=2018
and the value for Tr(A^-1)^2 - tr(A^-2) = 2/detA = 2/2018 = 1/1009

• たつき 9 months ago

Let A＝[a,b]
[c,d],
Let B be the inverse matrix of A,
then B＝[d/2018, −b/2018]
[−c/2018, a/2018].
Therefore, tr(B)²＝((a+d)/2018)²
We need the diagonal elements B²
(Not B² itself)
They are (d²+bc)/2018² and
(bc+a²)/2018²
Finally,
tr(B)²−tr(B²)
＝((a+d)²−(d²+bc)−(bc+a²))/2018²
＝2det(A)/2018²＝1/1009

• Nicholas Leclerc 9 months ago

The answer’s the inverse of 1009
For A = [a;b;c;d] {In the order: Top-left; Top-right; Bottom-left; Bottom-right}
Then there’s A^-1: [e;f;g;h], so that, for a vector v=[x;y]:
A(v)=[ax+cy;bx+dy]
Then A^-1(A(v))=
[e(ax+cy)+g(bx+dy);f(ax+cy)+h(bx+dy)]
=
[x;y]
Or:
x(ae+bg)+y(ce+dg)=x
x(af+bh)+y(cf+dh)=y
Thus:
ce+dg=0
ae+bg=1
cf+dh=1
af+bh=0
So:
-g(d/c)=(1-bg)/a
-h(b/a)=(1-dh)/c
So:
g(b/a-d/c)=1/a
h(d/c-b/a)=1/c
So all in all, A^-1=
Therefore A^-2=
THEREFORE, finally:
Or:
(2/2018^2)(2018)=1/1009
THEEEERE, Q-E-FREAKING-D 😰 phew...

• GreenMeansGO 9 months ago

Let A = [a, b; c, d] (representing standard matrix) Then ad-bc = 2018 and A_inverse is 1/2018*[d, -b; -c, a]. The trace squared is ((a+d)/2018)^2 = (a^2+2ad+d^2)/2018^2 (•)
The square of the inverse is 1/2018^2 * [d, -b; -c, a]*[d, -b; -c, a]
=1/2018^2 * [d^2 + bc, -ab-bd; -ac-cd, a^2+bc]. So the trace of the inverse squared is (d^2+bc+a^2+bc)/2018^2 = (a^2+2bc+d^2)/2018^2 (••).

• Xander Gouws 9 months ago

1/1009
Assume A = [a b, c d] and |A| = 2018
A^-1 = 1/|A| [d -b, -c a]
A^-2 = 1/(|A|^2) [d^2+bc -b(a+d), -c(a+d) a^2+bc]
Evaluating [tr(A^-1)]^2 - tr(A^-2) simplifies to 2(ad-bc)/(|A|^2) = 2/|A| = 2/2018 = 1/1009. The fact that the expression simplifies to a function of the determinant is enough proof to say that it is constant for all matrices with determinant 2018.

• Jack Miller 9 months ago

Let Matrix A = [ a b, c d ], and A⁻⁭¹ = det(A)⁻⁭¹ * [ d -b, -c a ].
Then write the initial set up as tr(det(A)⁻⁭¹ * [ d -b, -c a ])² - tr(det(A)⁻⁭² * [ d²+bc -db-ab, -cd-ac bc+a² ]) = C1
By removing the determinant by the argument because the trace function is linear, and squaring the first trace, you get:
((d+a)²-(a²+d²+2bc))/2018² = C1
Then multiply by 2018² to get another constant:
((d+a)²-(a²+d²+2bc)) = C2
By only removing the two ends of the second degree pascal’s triangle, you get:
2(ad-bc) = C2 {this is the determinant of a 2x2 matrix, and it was said to be 2018}
2(2018) = C2
C3 = C2
Q.E.D.
Whoops, I forgot to solve for C1.
C1 = 1/1009 {just keep the 2018² on the left}

• GreenMeansGO 9 months ago

I think you disqualified yourself by editing the comment.

• Benjamin Wang 9 months ago

Answer is 1/1009. Let A have rows a, b and c, d. The constant is (a+d)^2/2018-(a^2+d^2+2bc)/(2018^2) which after noting that detA=ad-bc in the second term, is -(-2detA)/(2018^2).

• Beshoy Nabil 9 months ago

Ans = 1/1009
Let A is [a b; c d] then determinant = ad-bc=2018
And A^-1 = 1/(ad-bc)*[d -c; -b a]
Also A^-2 = 1/(ad-bc)^2*[bc+d^2 -b*(a+d) ; -c*(a+d) bc+a^2]
Then tr(A^-2) = (a^2 + d^2 +2bc) / (ad-bc)^2

• SloppyTaco 9 months ago

I got 1/1009
I wrote a general matrix A of [{a,b},{c,d}] and wrote its inverse A^{-1}=[{d,-b},{-c,a}]. The trace of of the inverse is simply (a+d)/2018. The square of the inverse is [{d^2+bc,-bd-ba},{-cd-ca,bc+a^2}]/2018^2. The trace of the square of the inverse is (d^2+a^2+2bc)/2018^2. Taking the difference of the tr(A^{-1})^2 and tr(A^{-2}), you get: (a^2+2ad+d^2-a^2-d^2-2bc)/2018^2.
Simplifying this reduces to (2(ad-bc))/2018^2. The expression at the top is just the original A determinant, which was given as 2018. Therefore, (2*2018)/2018^2 = 2/2018 = 1/1009.

• Kevin Tong 9 months ago +3

Let the matrix be
[ a b ]
[ c d ]
Then the inverse if
1/2018 * [ d -b ]
[ -c a ]
Therefore, we have that (tr(A^-1))^2=((d+a)/2018)^2
Next, we notice that tr(A^-2)=(1/2018^2)*(d^2+bc+bc+a^2)=(1/2018^2)(d^2+2bc+a^2)
Therefore,

• Kevin Tong 9 months ago

Wow, first answer for once!!! and I didn't even need my notifications.

• Hung Hin Sun 9 months ago +6

By direct evaluation, the expression can be found to be 2/detA, which is a constant and is equal to 1/1009.

• Felix Trihardjo 9 months ago +1

Assume that A = [a b; c d], then det(A) = ad-bc = 2018 and inv(A) = [d -b; -c a]/2018. tr(inv(A))^2-tr(inv(A)^2) = (a+d)^2/2018^2-(d^2+2bc+a^2)/2018^2 = (2ad-2bc)/2018^2 = 2*det(A)/2018^2 = 1/1009
So tr(inv(A))^2-tr(inv(A)^2) = 1/1009.

• Kenny Yip 9 months ago +12

Let x, y be the eigenvalues of A.
det(A) = xy = 2018.
det(A inv) = 1/(xy) = 1/2018.
Tr(A inv)^2 - Tr(A inv^2) = (1/x + 1/y)^2 - (1/x^2 + 1/y^2) = 2/(xy) = 1/1009.
Here, I have used the properties that:
(1) determinant of a matrix is the product of the eigenvalues,
(2) trace of a matrix is the sum of the eigenvalues, and
(3) the eigenvalue of the inverse (square) of a non-singular matrix is the reciprocal (square) of each eigenvalue.

• • 