For what value of \(k\), the following system have a unique solution? (where R is set of real numbers)
\(x + y + z = 1\)
\(2x + 3y + 4z = 3\)
\(x - y + kz = 5\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(k \in R - \{-3\}\)
A system of three linear equations in three unknowns has a unique solution exactly when the determinant of the coefficient matrix is not zero.
\( \displaystyle |A| = \begin{vmatrix}1&1&1\\2&3&4\\1&-1&k\end{vmatrix} \)
Expanding along the first row:
\( \displaystyle |A| = 1\begin{vmatrix}3&4\\-1&k\end{vmatrix} - 1\begin{vmatrix}2&4\\1&k\end{vmatrix} + 1\begin{vmatrix}2&3\\1&-1\end{vmatrix} \)
\(= (3k + 4) - (2k - 4) + (-2 - 3)\)
\(= k + 3\)
So \(|A| \ne 0\) exactly when \(k \ne -3\): the system has a unique solution for every real k except \(-3\), that is, \(k \in R - \{-3\}\).