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
maganbhai P.
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#

Ratnaker Mehta
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!

#color(blue)("Enjoy Maths.!")#

Related questions
Impact of this question
5283 views around the world
You can reuse this answer
Creative Commons License