If findA^-1.

Hence, solve the system of equations:

3x + 3y+ 2z = 1

x + 2y = 4

2x – 3y –z = 5

OR

Find the inverse of the following matrix using elementary transformations.

|A| = 3(-2)-1(3) +2(-4)

= -6 -3 -8

= -17

∴ As, |A| is not equal to zero.

So, A^-1 exists.

Element of a co-factor matrix is given by,

C_ij = (-1)^i+j(M_ij)

Where M_ij is minor of A_ij.

M_ij is equal to the determinant of the matrix obtained after removing i^th row and j^th column.

So, C₁₁ = (-1)¹⁺¹

⇒ C₁₁ = -2 + 0

⇒ C₁₁ = -2

C₁₂ = (-1)¹⁺²

⇒ C₁₂ = -1(-3 + 6)

⇒ C₁₂ = -3

C₁₃ = (-1)¹⁺³

⇒ C₁₃ = 0 – 4

⇒ C₁₃ = -4

C₂₁ = (-1)²⁺¹

⇒ C₂₁ = (-1)(-1 + 0)

⇒ C₂₁ = 1

C₂₂ = (-1)²⁺²

⇒ C₂₂ = -3 – 4

⇒ C₂₂ = -7

C₂₃ = (-1)²⁺³

⇒ C₂₃ = (-1)(0 – (-2))

⇒ C₂₃ = (-1)(2)

⇒ C₂₃ = 2

C₃₁ = (-1)³⁺¹

⇒ C₃₁ = -3 – 4

⇒ C₃₁ = -7

C₃₂ = (-1)³⁺²

⇒ C₃₂ = (-1)(-9 – 6)

⇒ C₃₂ = (-1)(-15)

⇒ C₃₂ = 15

C₃₃ = (-1)³⁺³

⇒ C₃₃ = 6 – 3

⇒ C₃₃ = 3

So, co-factor matrix of A =

Now for given system of equations.

Clearly, (A^t)X = B

⇒ X = (A^t)^-1 B

As, (A^t)^-1 = (A^-1)^t

∴ X = (A^-1)^t B

Therefore, x = 2, y = 1, and z = –4

OR

Let,

As A = IA

So,

R₁→R₁+R₃

R₁→ (-1) R₁

R₂→ R₂ + 5R₁

R₃→ R₃ + 3R₁

Interchanging R₂ and R₃

R₂→ (-1) R₂

R₁→ R₁ + R₂

R₃→ R₃ + 2R₂

R₁→ R₁ - 9R₃

R₂→ R₂ - 152R₃

Therefore,