If the system of linear equations
\(x + 2y - 3z = 1\)
\((2p + 1)y + z = 2\)
\(3x + 3z = 5\)
has a unique solution, then p can not be equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{3}{4}\)
A system of three equations in three unknowns has a unique solution exactly when the determinant of the coefficients is non-zero.
\( \displaystyle \Delta = \begin{vmatrix} 1 & 2 & -3 \\ 0 & 2p+1 & 1 \\ 3 & 0 & 3 \end{vmatrix} \)
Expanding along the first row:
\(\Delta = 1\left[3(2p+1) - 0\right] - 2\left[0 - 3\right] - 3\left[0 - 3(2p+1)\right]\)
\(= 3(2p+1) + 6 + 9(2p+1)\)
\(= 24p + 18\)
For a unique solution \(\Delta \ne 0\):
\(24p + 18 \ne 0\)
\( \displaystyle p \ne -\frac{3}{4} \)