Using matrices, solve the following system of equations:

x + 2y - 3z = - 4

2x + 3y + 2z = 2

3x - 3y - 4z = 11

OR

If a, b, c is positive and unequal, show that the following determinant is negative:

Given: x + 2y - 3z = - 4, 2x + 3y + 2z = 2 and 3x - 3y - 4z = 11

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

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

AX = B where,

Now, check whether the system has a unique solution or not:

= 1{(-4)×(3) – 2×-3} – 2{(-4)×2 – (-3)×-3} + 3{2×2 – 3×-3}

= (-12 + 6) – 2(-8 – 9) + 3{4 + 9}

= -6 – 2(-17) + 3(13)

= -6 + 34 + 39

= 67

The system of the equation is consistent and have a unique solution

AX = B

⇒ X = A^-1 B

Formula used:

Thus,

X = A^-1 B

Hence, solutions of the equations are x = 3, y = -2, z = 1

OR

Given: a, b, c are positive

To prove: is negative

Applying C₁→ C₁ + C₂ + C₃

Applying R₂→ R₂ – R₁ and R₃→ R₃ – R₁

Expanding determinant by C₁

⇒ Δ = (a + b + c) [(c – b)(b – c) – (a – b)(a – c) – b(0) + c(0)]

⇒ Δ = (a + b + c) [(cb – b² – c² + bc) – (a² – ab – ac + bc)]

⇒ Δ = (a + b + c) (cb – b² – c² + bc – a² + ab + ac – bc)

⇒ Δ = (a + b + c) (– b² – c² + bc – a² + ab + ac)

⇒ Δ = -(a + b + c) (a² + b² + c² – bc – ab – ac)

Multiplying and dividing by 2:

Square of anything is always positive

⇒

We know,

a, b and c are positive

⇒ a + b + c >0

Therefore,

Hence Proved