To determine the number of real values of 'a' for which the given system of equations has nontrivial solutions, we need to consider the conditions for the system of linear equations to have nontrivial solutions. The system is:
This system of equations can be represented in matrix form as \(A \cdot X = 0\), where:
| \(A = \begin{pmatrix} 2 & 3 & a \\ 1 & a & -2 \\ 3 & 1 & 3 \end{pmatrix}\) | , | \(X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) |
For the system to have nontrivial solutions, the determinant of the coefficient matrix A must be zero: \(\det(A) = 0\).
Let's calculate the determinant of matrix A:
\(\det(A) = \begin{vmatrix} 2 & 3 & a \\ 1 & a & -2 \\ 3 & 1 & 3 \end{vmatrix}\)
Using the formula for a 3x3 determinant:
\(= 2 \begin{vmatrix} a & -2 \\ 1 & 3 \end{vmatrix} - 3 \begin{vmatrix} 1 & -2 \\ 3 & 3 \end{vmatrix} + a \begin{vmatrix} 1 & a \\ 3 & 1 \end{vmatrix}\)
\(= 2(a \cdot 3 - (-2) \cdot 1) - 3(1 \cdot 3 - (-2) \cdot 3) + a(1 \cdot 1 - 3 \cdot a)\)
\(= 2(3a + 2) - 3(3 + 6) + a(1 - 3a)\)
\(= 2(3a + 2) - 3 \times 9 + a - 3a^2\)
\(= 6a + 4 - 27 + a - 3a^2\)
\(= -3a^2 + 7a - 23\)
For nontrivial solutions, set the determinant to zero:
\(-3a^2 + 7a - 23 = 0\)
This is a quadratic equation in 'a'. For there to be real solutions to this equation, the discriminant must be non-negative. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\(\Delta = b^2 - 4ac\)
Here, \(a = -3\), \(b = 7\), \(c = -23\). Calculating the discriminant:
\(\Delta = 7^2 - 4(-3)(-23)\)
\(= 49 - 276\)
\(= -227\)
The discriminant is negative, indicating there are no real solutions for 'a'.
Therefore, the number of real values of 'a' for which the system of equations has nontrivial solutions is 0.