Answer :

**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 C_{1}→ C_{1} + C_{2} + C_{3}

Applying R_{2}→ R_{2} – R_{1} and R_{3}→ R_{3} – R_{1}

Expanding determinant by C_{1}

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

⇒ Δ = (a + b + c) [(cb – b^{2} – c^{2} + bc) – (a^{2} – ab – ac + bc)]

⇒ Δ = (a + b + c) (cb – b^{2} – c^{2} + bc – a^{2} + ab + ac – bc)

⇒ Δ = (a + b + c) (– b^{2} – c^{2} + bc – a^{2} + ab + ac)

⇒ Δ = -(a + b + c) (a^{2} + b^{2} + c^{2} – 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**

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Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Prove the followiMathematics - Board Papers