Q. 205.0( 2 Votes )

Solve the followi

Answer :

Given: - Equations are: –


x + y + z + w = 2


x – 2y + 2z + 2w = – 6


2x + y – 2z + 2w = – 5


3x – y + 3z – 3w = – 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 + z + w = 2


x – 2y + 2z + 2w = – 6


2x + y – 2z + 2w = – 5


3x – y + 3z – 3w = – 3


So by comparing with theorem, lets find D, D1, D2,D3 and D4



applying,



Solving determinant, expanding along 1st Row



applying,



D = 1[ – 6 – 88]


D = – 94


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


Here




applying,



Solving determinant, expanding along 1st Row



D1 = – 1{( – 10)[6( – 1) – 2( – 4)] – ( – 4)[( – 9)6 – ( – 4)3] + 0}


D1 = – 1{ – 10[ – 6 + 8] + 4[ – 54 + 12]}


D1 = – 1{ – 10[2] + 4[ – 42] }


D1 = 188


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


Here




applying,



Solving determinant, expanding along 1st Row



D2 = – 1{( – 1)[6( – 9) – 3( – 4)] – ( – 10)[0 – 6( – 4)] + 0[0 + 54]}


D2 = – 1{ – 1[ – 54 + 12] + 10(24) + 0}


D2 = – 282


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


Here




applying,



Solving determinant, expanding along 1st Row



D3 = – 1{( – 1)[ – 3 – ( – 9)2] – ( – 4)[0 – 6( – 9)] + ( – 10)[0 + 6]}


D3 = – 1{ – 1[15] + 4(54) – 10(6)}


D3 = – 1{ – 15 + 216 – 60}


D3 = – 141


And, Solve D4 formed by replacing 4th column by B matrices


Here




applying,



Solving determinant, expanding along 1st Row



D4 = ( – 3)[( – 9)( – 4) – 0] – 1[9 – ( – 4)( – 9)] + ( – 8)[0 – 16]


D4 = – 3[36] – 1(9 – 36) – 8( – 16)


D4 = – 108 + 27 + 128


D4 = 47


Thus by Cramer’s Rule, we have




x = – 2


again,




y = 3


again,





And,





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