Q. 185.0( 1 Vote )

Solve the followi

Answer :

Given: - Equations are: –


x + y = 1


x + z = – 6


x – y – 2z = 3


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 = 1


x + z = – 6


x – y – 2z = 3


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



Solving determinant, expanding along 1st Row


D = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)(1) – 1] + 0[ – 1 – 0]


D = 1[0 + 1] – 1[ – 3] – 0[ – 2]


D = 1 + 3 + 0


D = 4


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


Here




Solving determinant, expanding along 1st Row


D1 = 1[(0)( – 2) – (1)( – 1)] – 1[( – 2)( – 6) – 3] + 0[6 – 0]


D1 = 1[0 + 1] – 1[12 – 3] + 0[6]


D1 = 1[1] – 9 + 0


D1 = 1 – 9 + 0


D1 = – 8


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


Here




Solving determinant, expanding along 1st Row


D2 = 1[( – 6)( – 2) – (1)(3)] – 1[( – 2)(1) – 1] + 0[3 + 6]


D2 = 1[12 – 3] – 1( – 2 – 1) + 0(9)


D2 = 9 + 3


D2 = 12


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


Here




Solving determinant, expanding along 1st Row


D3 = 1[(0)(3) – ( – 1)( – 6)] – 1[(3)(1) – 1( – 6)] + 1[ – 1 + 0]


D3 = 1[0 – 6] – 1(3 + 6) + 1( – 1)


D3 = – 6 – 9 – 1


D3 = – 16


Thus by Cramer’s Rule, we have




x = – 2


again,




y = 3


and,




z = – 4


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