Q. 255.0( 2 Votes )

If A <span lang="

Answer :

Given, 3x + 4y + 7z = 14


2x – y + 3z = 4


x + 2y – 3z = 0


The above system can be written as:



AX = B


Now |A|= 3 (3-6) – 2(-12-14) +1(12+7)


= 3(-3)-2(-26) +1(19)


= -9 + 52 + 19


= 62


As |A|≠ 0,


So, A-1 exists.


Let Cij are cofactors of aij in A=[ aij]. Then,


C11 = (-1)1+1[3-6] =-3


C12 = (-1)1+2[-12-14] =26


C13 = (-1)1+3[12+7] =19


C21 = (-1)2+1[-6-3] =9


C22 = (-1)2+2[-9-7] =-16


C23 = (-1)2+3[9-14] =5


C31 = (-1)3+1[4+1] =5


C32 = (-1)3+2[6-4] =-2


C33 = (-1)1+1[-3-8] =-11


adj A =




Thus, the solution of system of equations is expressible as:



ATX = B


As |AT| = |A| = 62≠ 0


So, the solution can be given by:


X = (AT)-1B = (A-1)TB








Hence, x =1 , y = 1 and z = 1.


OR


We know that,


A=IA



R2️ R1



R2 R2 – 2R1




R3 R3 – 2R2




R1 R1 – R3




R2 R2 + R3




So,



As I=A-1A



Now we have to solve,








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