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?

Aug 7, 2018

Here ,
$L H S = | \left(1 , a , {a}^{2} , {a}^{3} + b c d\right) , \left(1 , b , {b}^{2} , {b}^{3} + c \mathrm{da}\right) , \left(1 , c , {c}^{2} , {c}^{3} + \mathrm{da} b\right) , \left(1 , d , {d}^{2} , {d}^{3} + a b c\right) |$

Using ${R}_{1} \leftrightarrow {C}_{1} , {R}_{2} \leftrightarrow {C}_{2} , {R}_{3} \leftrightarrow {C}_{3} \mathmr{and} {R}_{4} \leftrightarrow {C}_{4}$

$L H S = | \left(1 , 1 , 1 , 1\right) , \left(a , b , c , d\right) , \left({a}^{2} , {b}^{2} , {c}^{2} , {d}^{2}\right) , \left({a}^{3} + b c d , {b}^{3} + c \mathrm{da} , {c}^{3} + \mathrm{da} b , {d}^{3} + a b c\right) |$

Let ,$P = {a}^{3} + b c d , Q = {b}^{3} + c \mathrm{da} , R = {c}^{3} + \mathrm{da} b , S = {d}^{3} + a b c$

and taking ${C}_{1} - {C}_{2} , {C}_{2} - {C}_{3} , {C}_{3} - {C}_{4}$

$L H S = | \left(0 , 0 , 0 , 1\right) , \left(a - b , b - c , c - d , d\right) , \left({a}^{2} - {b}^{2} , {b}^{2} - {c}^{2} , {c}^{2} - {d}^{2} , {d}^{2}\right) , \left(P - Q , Q - R , R - S , S\right) |$

Simplifying and reducing the determinant for $1$

$L H S = 1 \cdot | \left(a - b , b - c , c - d\right) , \left({a}^{2} - {b}^{2} , {b}^{2} - {c}^{2} , {c}^{2} - {d}^{2}\right) , \left(P - Q , Q - R , R - S\right) |$
Now,

$P - Q = {a}^{3} + b c d - {b}^{3} - c \mathrm{da} = {a}^{3} - {b}^{3} - c \mathrm{da} + b c d$

$P - Q = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) - c d \left(a - b\right)$

$P - Q = \left(a - b\right) \left({a}^{2} + {b}^{2} + a b - c d\right)$

Similarly,

$Q - R = \left(b - c\right) \left({b}^{2} + {c}^{2} + b c - a d\right)$

$R - S = \left(c - d\right) \left({c}^{2} + {d}^{2} + c d - a b\right)$

Subst.values of $P - Q , Q - R , R - S \mathmr{and} t h e n$

taking ${C}_{1} \left(\frac{1}{a - b}\right) , {C}_{2} \left(\frac{1}{b - c}\right) \mathmr{and} {C}_{3} \left(\frac{1}{c - d}\right)$

$L H S = \left(a - b\right) \left(b - c\right) \left(c - d\right) \times | \left(1 , 1 , 1\right) , \left(a + b , b + c , c + d\right) , \left({a}^{2} + {b}^{2} + a b - c d , {b}^{2} + {c}^{2} + b c - a d , {c}^{2} + {d}^{2} + c d - a b\right) |$

Let ,

$L = {a}^{2} + {b}^{2} + a b - c d ,$

$M = {b}^{2} + {c}^{2} + b c - a d ,$

$N = {c}^{2} + {d}^{2} + c d - a b$

Taking ${C}_{1} - {C}_{2} \mathmr{and} {C}_{2} - {C}_{3}$

$L H S = \left(a - b\right) \left(b - c\right) \left(c - d\right) \times | \left(0 , 0 , 1\right) , \left(a - c , b - d , c + d\right) , \left(L - M , M - N , N\right) |$
Now,

$L - M = {a}^{2} - {c}^{2} + a b - b c + a d - c d$

$L - M = \left(a - c\right) \left(a + b + c + d\right)$

similarly ,

$M - N = \left(b - d\right) \left(a + b + c + d\right)$

Subst. values of $L - M \mathmr{and} M - N$,then

taking ${C}_{1} \left(\frac{1}{a - c}\right) \mathmr{and} {C}_{2} \left(\frac{1}{b - d}\right)$

$L H S = \left(a - b\right) \left(b - c\right) \left(c - d\right) \left(a - c\right) \left(b - d\right) \times | \left(0 , 0 , 1\right) , \left(1 , 1 , c + d\right) , \left(a + b + c + d , a + b + c + d , N\right) |$

Using property:$\left[{C}_{1} = {C}_{2}\right] \implies D = 0$

$L H S = \left(a - b\right) \left(b - c\right) \left(c - d\right) \left(a - c\right) \left(b - d\right) \times \left\{0\right\}$

$\therefore L H S = 0 = R H S$

Aug 7, 2018

Please refer to an Aliter in The Explanation.

Explanation:

Let, $D = | \left(1 , a , {a}^{2} , p\right) , \left(1 , b , {b}^{2} , q\right) , \left(1 , c , {c}^{2} , r\right) , \left(1 , d , {d}^{2} , s\right) |$,

where, $p = {a}^{3} + b c d , q = {b}^{2} + c \mathrm{da} , r = {c}^{3} + \mathrm{da} b , s = {d}^{3} + a b c$.

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

${D}_{1} \mathmr{and} {D}_{2}$, where,

${D}_{1} = | \left(1 , a , {a}^{2} , {a}^{3}\right) , \left(1 , b , {b}^{2} , {b}^{3}\right) , \left(1 , c , {c}^{2} , {c}^{3}\right) , \left(1 , d , {d}^{2} , {d}^{3}\right) | ,$ &,

${D}_{2} = | \left(1 , a , {a}^{2} , b c d\right) , \left(1 , b , {b}^{2} , c \mathrm{da}\right) , \left(1 , c , {c}^{2} , \mathrm{da} b\right) , \left(1 , d , {d}^{2} , a b c\right) | .$

So, $D = {D}_{1} + {D}_{2}$.

By ${R}_{1} \times a , {R}_{2} \times b , {R}_{3} \times c , {R}_{4} \times d ,$ we have,

$\left(a b c d\right) {D}_{2} = | \left(a , {a}^{2} , {a}^{3} , a b c d\right) , \left(b , {b}^{2} , {b}^{3} , b c \mathrm{da}\right) , \left(c , {c}^{2} , {c}^{3} , c \mathrm{da} b\right) , \left(d , {d}^{2} , {d}^{3} , \mathrm{da} b c\right) | , \mathmr{and} ,$

${D}_{2} = | \left(a , {a}^{2} , {a}^{3} , 1\right) , \left(b , {b}^{2} , {b}^{3} , 1\right) , \left(c , {c}^{2} , {c}^{3} , 1\right) , \left(d , {d}^{2} , {d}^{3} , 1\right) | .$

Next, the interchange between ${C}_{1} \text{ & } {C}_{4}$ will give $- {D}_{2} ,$

$i . e . , - {D}_{2} = | \left(1 , {a}^{2} , {a}^{3} , a\right) , \left(1 , {b}^{2} , {b}^{3} , b\right) , \left(1 , {c}^{2} , {c}^{3} , c\right) , \left(1 , {d}^{2} , {d}^{3} , d\right) | .$

Similarly, ${C}_{2} \leftrightarrow {C}_{4}$ will yield $- \left(- {D}_{2}\right) = {D}_{2}$.

$\therefore {D}_{2} = | \left(1 , a , {a}^{3} , {a}^{2}\right) , \left(1 , b , {b}^{3} , {b}^{2}\right) , \left(1 , c , {c}^{3} , {c}^{2}\right) , \left(1 , d , {d}^{3} , {d}^{2}\right) | .$

Finally, ${C}_{3} \leftrightarrow {C}_{4}$ will result in $- {D}_{2}$.

$\therefore - {D}_{2} = | \left(1 , a , {a}^{2} , {a}^{3}\right) , \left(1 , b , {b}^{2} , {b}^{3}\right) , \left(1 , c , {c}^{2} , {c}^{3}\right) , \left(1 , d , {d}^{2} , {d}^{3}\right) | .$

But then, this means that, $- {D}_{2} = {D}_{1}$, or,

${D}_{1} + {D}_{2} = D = 0$, as Respected Maganbhai P. has readily

shown!

$\textcolor{b l u e}{\text{Enjoy Maths.!}}$