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 C_{ij} are cofactors of a_{ij} in A=[ a_{ij}]. Then,

C_{11} = (-1)^{1+1}[3-6] =-3

C_{12} = (-1)^{1+2}[-12-14] =26

C_{13} = (-1)^{1+3}[12+7] =19

C_{21} = (-1)^{2+1}[-6-3] =9

C_{22} = (-1)^{2+2}[-9-7] =-16

C_{23} = (-1)^{2+3}[9-14] =5

C_{31} = (-1)^{3+1}[4+1] =5

C_{32} = (-1)^{3+2}[6-4] =-2

C_{33} = (-1)^{1+1}[-3-8] =-11

∴ adj A =

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

⇒ A^{T}X = B

As |A^{T}| = |A| = 62≠ 0

So, the solution can be given by:

X = (A^{T})^{-1}B = (A^{-1})^{T}B

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

**OR**

We know that,

A=IA

R_{2}↔️ R_{1}

R_{2} → R_{2} – 2R_{1}

R_{3} → R_{3} – 2R_{2}

R_{1} → R_{1} – R_{3}

R_{2} → R_{2} + R_{3}

So,

As I=A^{-1}A

Now we have to solve,

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