Q. 195.0( 2 Votes )

Solve the followi

Answer :

Given: - Equations are: –


x + y + z + 1 = 0


ax + by + cz + d = 0


a2x + b2y + c2z + d2 = 0


Tip: - Theorem – Cramer’s Rule


Let there be a system of n simultaneous linear equations and with n unknown given by







and let Dj be the determinant obtained from D after replacing the jth column by



Then,


provided that D ≠ 0


Now, here we have


x + y + z + 1 = 0


ax + by + cz + d = 0


a2x + b2y + c2z + d2 = 0


So by comparing with theorem, lets find D , D1 , D2 and D3



applying,



Take (b – a) from c2 , and (c – a) from c3 common, we get



Solving determinant, expanding along 1st Row


D = (b – a)(c – a)1[c + a – (b + a)]


D = (b – a)(c – a)(c + a – b – a)


D = (b – a)(c – a)(c – b)


D = (a – b)(b – c)(c – a)


Again, Solve D1 formed by replacing 1st column by B matrices


Here




applying,



Take (b – d) from c2 , and (c – d) from c3 common, we get



Solving determinant, expanding along 1st Row


D1 = – (b – d)(c – d)1[c + d – (b + d)]


D1 = – (b – d)(c – d)(c + d – b – d)


D1 = – (b – d)(c – d)(c – b)


D1 = – (d – b)(b – c)(c – d)


Again, Solve D2 formed by replacing 2nd column by B matrices


Here




applying,



Take (d – a) from c2 , and (c – a) from c3 common, we get



Solving determinant, expanding along 1st Row


D2 = – (d – a)(c – a)1[c + a – (d + a)]


D2 = – (d – a)(c – a)(c + a – d – a)


D2 = – (d – a)(c – a)(c – d)


D2 = – (a – d)(d – c)(c – a)


And, Solve D3 formed by replacing 3rd column by B matrices


Here




applying,



Take (b – a) from c2 , and (d – a) from c3 common, we get



Solving determinant, expanding along 1st Row


D3 = – (b – d)(c – d)1[a + d – (b + a)]


D3 = – (b – d)(c – d)(a + d – b – a)


D3 = – (b – d)(c – d)(d – b)


D3 = – (d – b)(b – d)(c – d)


Thus by Cramer’s Rule, we have




again,




and,




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