Prove that is divisible by and hence find the quotient.

OR

Using elementary transformations, find the inverse of the matrix and use it to solve the following system of linear equations:

To prove: is divisible by (x + y + z)

To find: the quotient

Apply C₁→ C₁ – C₂ and C₂→ C₂ – C₃

Taking common (x + y + z) from C₁ and C₂

Apply R₁→ R₁ + R₂ + R₃

Expanding the determinant

= (x + y + z)² (xy – z² + yz – x² + zx – y²){(z – y)(y – x) – (x – z)²}

= (x + y + z)² (xy – z² + yz – x² + zx – y²){(zy – y² + xy – xz) – (x² + z² – 2xz)}

= (x + y + z)² (xy – z² + yz – x² + zx – y²)(zy – y² + xy – xz – x² – z² + 2xz)

= (x + y + z)² (zy + xy + xz – x² – y² – z²)²

Hence the given determinant is divisible by (x + y + z) and quotient is

OR

Given: System of equations:

8x + 4y + 3z = 19, 2x + y + z = 5, x + 2y + 2z = 7

To find: Solution of the system of the equations i.e. values of x, y and z which satisfy these equations

We know,

A = I.A where I is an identity matrix

Applying R₁ R₃

Applying R₂ → R₂ – 2R₁ and R₃ → R₃ – 8R₁

Applying R₁ → R₁ – 2R₂ and R₃ → R₃ + 12R₂

Applying R₃ → -R₃ and R₂ → R₂ – R₃

To solve these equation and get values of x, y and z, we have:

AX = B where,

AX = B

⇒ X = A^-1 B

Hence, solutions of the equations are x = 1, y = 2, z = 1