Q. 235.0( 1 Vote )

Using matrices, s

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

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