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For some constant 'k', if the system of linear equations

2x−y+3z=12x - y + 3z = 1

x−2y+z=3x - 2y + z = 3

kx+y−z=0kx + y - z = 0

has a unique solution, then

Solution

✅ Correct Option: 4

For a system of equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

The coefficient matrix from the three equations is:

A=(2−131−21k1−1)A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & -2 & 1 \\ k & 1 & -1 \end{pmatrix}


Using first row expansion to calculate the determinant:

det⁡(A)=2∣−211−1∣−(−1)∣11k−1∣+3∣1−2k1∣\det(A) = 2\begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} - (-1)\begin{vmatrix} 1 & 1 \\ k & -1 \end{vmatrix} + 3\begin{vmatrix} 1 & -2 \\ k & 1 \end{vmatrix}

For the first term:

2[(−2)(−1)−(1)(1)]2[(-2)(-1) - (1)(1)]

=2[2−1]= 2[2 - 1]

=2= 2

For the second term:

+1[(1)(−1)−(1)(k)]+1[(1)(-1) - (1)(k)]

=−1−k= -1 - k

For the third term:

3[(1)(1)−(−2)(k)]3[(1)(1) - (-2)(k)]

=3[1+2k]= 3[1 + 2k]

=3+6k= 3 + 6k


Combining all terms:

det⁡(A)=2+(−1−k)+(3+6k)\det(A) = 2 + (-1 - k) + (3 + 6k)

=2−1−k+3+6k= 2 - 1 - k + 3 + 6k

=4+5k= 4 + 5k


For a unique solution:

det⁡(A)≠0\det(A) \neq 0

4+5k≠04 + 5k \neq 0

5k≠−45k \neq -4

k≠−45k \neq -\dfrac{4}{5}

Therefore, the system has a unique solution when k≠−45k \neq -\dfrac{4}{5}.

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