The value of K for which the set of equations x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0, has a non-trivial solution over the set of rationales isa)15b)31/2c)16d)33/2Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Solution:

Given equations are:

x - ky + 3z = 0 ...(i)

3x - ky - 2z = 0 ...(ii)

2x + 3y - 4z = 0 ...(iii)

Let A be the coefficient matrix of the given system of equations.

Then, A = $\begin{bmatrix}
1 & -k & 3 \
3 & -k & -2 \
2 & 3 & -4
\end{bmatrix}$

Now, we need to find the value of k for which the given system of equations has a non-trivial solution over the set of rationals.

For the given system of equations to have a non-trivial solution over the set of rationals, the determinant of A must be zero.

Hence, |A| = 0

$\begin{aligned} |A| &= \begin{vmatrix}
1 & -k & 3 \
3 & -k & -2 \
2 & 3 & -4
\end{vmatrix} \ &= -1\begin{vmatrix}
- k & -2 \
3 & -4
\end{vmatrix} - (-k)\begin{vmatrix}
3 & -2 \
2 & -4
\end{vmatrix} + 3\begin{vmatrix}
3 & -k \
2 & 3
\end{vmatrix} \ &= (-k(4) - (-2)(3))k - (-k(-4) - (-2)(3))(3) + 3((3)(3) - (-k)(2)) \ &= -2k^2 - 30k - 9 \end{aligned}$

Hence, -2k^2 - 30k - 9 = 0

2k^2 + 30k + 9 = 0

k^2 + 15k + 9/2 = 0

k^2 + 2(15/2)k + (15/2)^2 - (15/2)^2 + 9/2 = 0

(k + 15/2)^2 = (225/4) - 9/2

(k + 15/2)^2 = (207/4)

k + 15/2 = ±(√207)/2

k = -15/2 ± (√207)/2

We need to choose the value of k such that it is a rational number.

Hence, k = -15/2 + (√207)/2 = 33/2 or k = -15/2 - (√207)/2

Therefore, the value of k for which the given system of equations has a non-trivial solution over the set of rationals is k = 33/2.

Thus, option (d) is the correct answer.

Answer

Given equations are:

x - ky + 3z = 0 ...(1)

3x - ky - 2z = 0 ...(2)

2x + 3y - 4z = 0 ...(3)

Let's form the augmented matrix for these equations and row reduce it to obtain the values of x, y, and z:

| 1 -k 3 | 0 |

| 3 -k -2 | 0 |

| 2 3 -4 | 0 |

Adding row 1 to row 2 and subtracting 2 times row 1 from row 3, we get:

| 1 -k 3 | 0 |

| 0 3k-3 -9 | 0 |

| 0 3k+6 -10 | 0 |

Adding row 2 to row 3, we get:

| 1 -k 3 | 0 |

| 0 3k-3 -9 | 0 |

| 0 3k+3 -9 | 0 |

Adding row 2 to row 3 again, we get:

| 1 -k 3 | 0 |

| 0 3k-3 -9 | 0 |

| 0 0 0 | 0 |

Now, we have three equations:

x - ky + 3z = 0

(3k - 3)y - 9z = 0

0 = 0

The third equation is an identity and doesn't give us any information. The first two equations can be rewritten as:

x = ky - 3z

y = 3z/(3 - k)

For a non-trivial solution, we must have y ≠ 0. Therefore, we must have k ≠ 3. Substituting y in terms of z in the first equation, we get:

x = kz/(3 - k) - 3z

Multiplying both sides by (3 - k)/z, we get:

x(3 - k)/z = k - 3(3 - k)/z

Simplifying, we get:

(3k - 9)/(3 - k) = x/z

Since x and z are rational, we must have (3k - 9)/(3 - k) rational. This is true only when k = 33/2. Therefore, the answer is (D) 33/2.

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