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Question: If x, y, z ∈ R & $\Delta = \begin{vmatrix} x & x+y & x+y+z \\ 2x & 5x+2y & 7x+5y+2z \\ 3x & 7x+3y & ...

If x, y, z ∈ R & Δ=∣xx+yx+y+z2x5x+2y7x+5y+2z3x7x+3y9x+7y+3z∣=−16\Delta = \begin{vmatrix} x & x+y & x+y+z \\ 2x & 5x+2y & 7x+5y+2z \\ 3x & 7x+3y & 9x+7y+3z \end{vmatrix} = -16 then value of x is

A

-2

B

-3

C

2

D

3

Answer

2

Explanation

Solution

The given determinant is Δ=∣xx+yx+y+z2x5x+2y7x+5y+2z3x7x+3y9x+7y+3z∣\Delta = \begin{vmatrix} x & x+y & x+y+z \\ 2x & 5x+2y & 7x+5y+2z \\ 3x & 7x+3y & 9x+7y+3z \end{vmatrix}. We are given that Δ=−16\Delta = -16.

We can simplify the determinant using column operations. Apply the operation C2←C2−C1C_2 \leftarrow C_2 - C_1:

Δ=∣x(x+y)−xx+y+z2x(5x+2y)−2x7x+5y+2z3x(7x+3y)−3x9x+7y+3z∣=∣xyx+y+z2x3x+2y7x+5y+2z3x4x+3y9x+7y+3z∣\Delta = \begin{vmatrix} x & (x+y)-x & x+y+z \\ 2x & (5x+2y)-2x & 7x+5y+2z \\ 3x & (7x+3y)-3x & 9x+7y+3z \end{vmatrix} = \begin{vmatrix} x & y & x+y+z \\ 2x & 3x+2y & 7x+5y+2z \\ 3x & 4x+3y & 9x+7y+3z \end{vmatrix}

Apply the operation C3←C3−C2C_3 \leftarrow C_3 - C_2:

Δ=∣xy(x+y+z)−(x+y)2x3x+2y(7x+5y+2z)−(5x+2y)3x4x+3y(9x+7y+3z)−(7x+3y)∣=∣xyz2x3x+2y2x+3y+2z3x4x+3y2x+4y+3z∣\Delta = \begin{vmatrix} x & y & (x+y+z)-(x+y) \\ 2x & 3x+2y & (7x+5y+2z)-(5x+2y) \\ 3x & 4x+3y & (9x+7y+3z)-(7x+3y) \end{vmatrix} = \begin{vmatrix} x & y & z \\ 2x & 3x+2y & 2x+3y+2z \\ 3x & 4x+3y & 2x+4y+3z \end{vmatrix}

Now, apply row operations to simplify the determinant further. Apply the operation R2←R2−2R1R_2 \leftarrow R_2 - 2R_1: The new R2R_2 is (2x−2x,(3x+2y)−2y,(2x+3y+2z)−2z)=(0,3x,2x+3y)(2x-2x, (3x+2y)-2y, (2x+3y+2z)-2z) = (0, 3x, 2x+3y).

Apply the operation R3←R3−3R1R_3 \leftarrow R_3 - 3R_1: The new R3R_3 is (3x−3x,(4x+3y)−3y,(2x+4y+3z)−3z)=(0,4x,2x+4y)(3x-3x, (4x+3y)-3y, (2x+4y+3z)-3z) = (0, 4x, 2x+4y).

The determinant becomes: Δ=∣xyz03x2x+3y04x2x+4y∣\Delta = \begin{vmatrix} x & y & z \\ 0 & 3x & 2x+3y \\ 0 & 4x & 2x+4y \end{vmatrix}

Expand the determinant along the first column:

Δ=x⋅∣3x2x+3y4x2x+4y∣−0⋅∣yz4x2x+4y∣+0⋅∣yz3x2x+3y∣\Delta = x \cdot \begin{vmatrix} 3x & 2x+3y \\ 4x & 2x+4y \end{vmatrix} - 0 \cdot \begin{vmatrix} y & z \\ 4x & 2x+4y \end{vmatrix} + 0 \cdot \begin{vmatrix} y & z \\ 3x & 2x+3y \end{vmatrix}

Δ=x[(3x)(2x+4y)−(4x)(2x+3y)]\Delta = x [ (3x)(2x+4y) - (4x)(2x+3y) ]

Δ=x[(6x2+12xy)−(8x2+12xy)]\Delta = x [ (6x^2 + 12xy) - (8x^2 + 12xy) ]

Δ=x[6x2+12xy−8x2−12xy]\Delta = x [ 6x^2 + 12xy - 8x^2 - 12xy ]

Δ=x[−2x2]\Delta = x [ -2x^2 ]

Δ=−2x3\Delta = -2x^3

We are given that Δ=−16\Delta = -16. So, −2x3=−16-2x^3 = -16. Divide both sides by -2: x3=−16−2x^3 = \frac{-16}{-2}

x3=8x^3 = 8

Since x∈Rx \in \mathbb{R}, the only real solution is the cube root of 8. x=83x = \sqrt[3]{8}

x=2x = 2

The value of x is 2.