91Ó°ÊÓ

We will represent the given system of equations in the form of an augmented matrix as follows: \( \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & 1 & 7 \\ 1 & 1 & -k^2 & -k \end{array} \right] \)

We will now perform Gaussian Elimination to simplify the matrix: First operation: Subtract row 1 from row 3: \( \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & 1 & 7 \\ 0 & -1 & -k^2+1 & -k-3 \end{array} \right] \) Second operation: Subtract twice row 1 from row 2: \( \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -k^2+1 & -k-3 \end{array} \right] \) Third operation: Add row 2 to row 3: \( \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & -k^2+4 & -k-2 \end{array} \right] \)

(a) No solution: For the system to have no solution, we need the last row of the matrix to be in the form \(0x_1 + 0x_2 + 0x_3 = c\) where \(c \neq 0\). This means that \(-k^2 + 4 = 0\) and \(-k - 2 \neq 0\). Solving for k, we get \(k = \pm 2\). However, when \(k = 2\), \(-k - 2 = -4\) which is not 0, so the system has no solution only when \(k = -2\). (b) Infinite number of solutions: For the system to have an infinite number of solutions, we need the last row of the matrix to be in the form \(0x_1 + 0x_2 + 0x_3 = 0\). This means that \(-k^2 + 4 = 0\) and \(-k - 2 = 0\). Solving for k, we again get \(k = \pm 2\). However, this time, when \(k = 2\), \(-k - 2 = 0\) which satisfies the condition. So the system has an infinite number of solutions when \(k = 2\). (c) Unique solution: For the system to have a unique solution, the matrix should be in triangular form with non-zero coefficients on the diagonal, which means \(-k^2 + 4 \neq 0\). In this case, the value of k should be different from the values obtained in the above cases. Therefore, the system has a unique solution when \(k \neq 2, -2\).

The constant k has the following values for each case: (a) No solution: \(k = -2\) (b) Infinite number of solutions: \(k = 2\) (c) Unique solution: \(k \neq 2, -2\)