# If A <span lang="

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,      Rate this question :

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