Prove |(1,a,a^2,a^3+bcd),(1,b,b^2,b^3+cda),(1,c,c^2,c^3+dab),(1,d,d^2,d^3+abc)|=0?

2 Answers
Aug 7, 2018

Here ,
LHS=|(1,a,a^2,a^3+bcd),(1,b,b^2,b^3+cda),(1,c,c^2,c^3+dab),(1,d,d^2,d^3+abc)|

Using R_1harrC_1, R_2harrC_2 ,R_3harrC_3and R_4harrC_4

LHS=|(1,1,1,1),(a,b,c,d),(a^2,b^2,c^2,d^2),(a^3+bcd,b^3+cda,c^3+dab,d^3+abc)|

Let ,P=a^3+bcd, Q=b^3+cda, R=c^3+dab,S=d^3+abc

and taking C_1-C_2, C_2-C_3 ,C_3-C_4

LHS=|(0,0,0,1),(a-b,b-c,c-d,d),(a^2-b^2,b^2-c^2,c^2-d^2,d^2),(P-Q,Q-R,R-S,S)|

Simplifying and reducing the determinant for 1

LHS=1*|(a-b,b-c,c-d),(a^2-b^2,b^2-c^2,c^2-d^2),(P-Q,Q-R,R-S)|
Now,

P-Q=a^3+bcd-b^3-cda=a^3-b^3-cda+bcd

P-Q=(a-b)(a^2+ab+b^2)-cd(a-b)

P-Q=(a-b)(a^2+b^2+ab-cd)

Similarly,

Q-R=(b-c)(b^2+c^2+bc-ad)

R-S=(c-d)(c^2+d^2+cd-ab)

Subst.values of P-Q,Q-R,R-S and then

taking C_1(1/(a-b)),C_2(1/(b-c)) and C_3(1/(c-d))

LHS=(a-b)(b-c)(c-d) xx|(1,1,1),(a+b,b+c,c+d),(a^2+b^2+ab-cd,b^2+c^2+bc-ad,c^2+d^2+cd-ab)|

Let ,

L=a^2+b^2+ab-cd,

M=b^2+c^2+bc-ad,

N=c^2+d^2+cd-ab

Taking C_1-C_2 and C_2-C_3

LHS=(a-b)(b-c)(c-d) xx|(0,0,1),(a-c,b-d,c+d),(L-M,M-N,N)|
Now,

L-M=a^2-c^2+ab-bc+ad-cd

L-M=(a-c)(a+b+c+d)

similarly ,

M-N=(b-d)(a+b+c+d)

Subst. values of L-M andM-N,then

taking C_1(1/(a-c)) and C_2(1/(b-d))

LHS=(a-b)(b-c)(c-d)(a-c)(b-d) xx|(0,0,1),(1,1,c+d),(a+b+c+d,a+b+c+d,N)|

Using property:[C_1=C_2]=>D=0

LHS=(a-b)(b-c)(c-d)(a-c)(b-d) xx{0}

:.LHS=0=RHS

Aug 7, 2018

Please refer to an Aliter in The Explanation.

Explanation:

Let, D=|(1,a,a^2,p),(1,b,b^2,q),(1,c,c^2,r),(1,d,d^2,s)|,

where, p=a^3+bcd, q=b^2+cda, r=c^3+dab, s=d^3+abc.

Note that, D can be split as the sum of two determinants

D_1 and D_2, where,

D_1=|(1,a,a^2,a^3),(1,b,b^2,b^3),(1,c,c^2,c^3),(1,d,d^2,d^3)|, &,

D_2=|(1,a,a^2,bcd),(1,b,b^2,cda),(1,c,c^2,dab),(1,d,d^2,abc)|.

So, D=D_1+D_2.

By R_1xxa, R_2xxb, R_3xxc, R_4xxd, we have,

(abcd)D_2=|(a,a^2,a^3,abcd),(b,b^2,b^3,bcda),(c,c^2,c^3,cdab),(d,d^2,d^3,dabc)|, or,

D_2=|(a,a^2,a^3,1),(b,b^2,b^3,1),(c,c^2,c^3,1),(d,d^2,d^3,1)|.

Next, the interchange between C_1" & "C_4 will give -D_2,

i.e., -D_2=|(1,a^2,a^3,a),(1,b^2,b^3,b),(1,c^2,c^3,c),(1,d^2,d^3,d)|.

Similarly, C_2harr C_4 will yield -(-D_2)=D_2.

:.D_2=|(1,a,a^3,a^2),(1,b,b^3,b^2),(1,c,c^3,c^2),(1,d,d^3,d^2)|.

Finally, C_3harrC_4 will result in -D_2.

:.-D_2=|(1,a,a^2,a^3),(1,b,b^2,b^3),(1,c,c^2,c^3),(1,d,d^2,d^3)|.

But then, this means that, -D_2=D_1, or,

D_1+D_2=D=0, as Respected Maganbhai P. has readily

shown!

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