The value of \( a \) for which the system \[ ax + y + z = 0,\; x + ay + z = 0,\; x + y + z = 0 \] has non-zero solutions is
To determine the value of \( a \) for which the system of linear equations has non-zero solutions, we need to analyze the given system of equations:
\[\begin{align*} ax + y + z &= 0, \\ x + ay + z &= 0, \\ x + y + z &= 0. \end{align*}\]The system of equations can be expressed in matrix form as follows:
| \[\begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\] |
For the system to have non-zero solutions, the determinant of the coefficient matrix must be zero. The determinant can be calculated as follows:
\[\text{det}\left(\begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & 1 \end{bmatrix}\right) = a \begin{vmatrix} a & 1 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & a \\ 1 & 1 \end{vmatrix}\]Compute each 2x2 determinant individually:
\[\begin{vmatrix} a & 1 \\ 1 & 1 \end{vmatrix} = a(1) - 1(1) = a - 1\]\[\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1\cdot1 - 1\cdot1 = 0\]\[\begin{vmatrix} 1 & a \\ 1 & 1 \end{vmatrix} = 1(1) - a(1) = 1 - a\]Substitute back into the determinant expression:
\[\text{det} = a(a - 1) - 0 + (1 - a) = a(a - 1) + 1 - a\]Simplifying further:
\[= a^2 - a + 1 - a = a^2 - 2a + 1 = (a - 1)^2\]For the system to have non-zero solutions, set the determinant equal to zero:
\[(a - 1)^2 = 0\]Solve for \( a \):
The solution is \( a = 1 \).
Therefore, the correct answer is: 1