If A , then find A^-1 and hence solve the following system of equations:

3x + 4y + 7z = 14, 2x – y + 3z = 4, x + 2y – 3z = 0

OR

If , find the inverse of A using elementary row transformations and hence solve the following matrix equation

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,

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:

⇒ A^TX = B

As |A^T| = |A| = 62≠ 0

So, the solution can be given by:

X = (A^T)^-1B = (A^-1)^TB

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

OR

We know that,

A=IA

R₂↔️ R₁

R₂ → R₂ – 2R₁

R₃ → R₃ – 2R₂

R₁ → R₁ – R₃

R₂ → R₂ + R₃

So,

As I=A^-1A

Now we have to solve,