NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

The value of $k$ for which the set of equations $3x + ky - 2z = 0$ , $x + ky + 3z = 0$ and $2x + 3y - 4z = 0$ has a non-trivial solution, is:

Choose the correct answer:

Explanation

1. Write the determinant of the coefficients:

The system is:

3x + ky - 2z = 0

x + ky + 3z = 0

2x + 3y - 4z = 0

The determinant $\Delta$ is formed by the coefficients of $x, y,$ and $z$ :

\Delta = \begin{vmatrix} 3 & k & -2 \\ 1 & k & 3 \\ 2 & 3 & -4 \end{vmatrix} = 0

2. Expand the determinant:

Expanding along the first row:

3[k(-4) - 3(3)] - k[1(-4) - 2(3)] + (-2)[1(3) - 2(k)] = 0

Now, simplify each term:

3[-4k - 9] - k[-4 - 6] - 2[3 - 2k] = 0

-12k - 27 - k[-10] - 6 + 4k = 0

-12k - 27 + 10k - 6 + 4k = 0

3. Solve for $k$ :

Combine the $k$ terms and the constant terms:

(-12k + 10k + 4k) + (-27 - 6) = 0

2k - 33 = 0

2k = 33

k = \frac{33}{2}

Conclusion:

For the system to have a non-trivial solution, the value of $k$ must be $\frac{33}{2}$ .

Correct Option:

(d) $\frac{33}{2}$