Answer :
To determine if the given system of homogeneous linear equations has a non-trivial solution, we need to analyze the determinant of its coefficient matrix. Here are the detailed steps:
1. Form the Coefficient Matrix:
The coefficient matrix \(A\) for the given system of equations can be written as:
[tex]\[ A = \begin{bmatrix} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9 \end{bmatrix} \][/tex]
2. Calculate the Determinant of the Matrix \(A\):
The determinant of a 3x3 matrix:
[tex]\[ \text{det}(A) = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \][/tex]
is calculated using the formula:
[tex]\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]
For our matrix \(A\):
[tex]\[ A = \begin{bmatrix} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9 \end{bmatrix} \][/tex]
Applying the formula, the determinant \(\text{det}(A)\) is:
[tex]\[ \text{det}(A) = 1(4 \cdot 9 - (-8) \cdot (-7)) - 3(1 \cdot 9 - (-8) \cdot (-3)) + (-5)(1 \cdot (-7) - 4 \cdot (-3)) \][/tex]
Simplifying the expression, we get:
[tex]\[ \text{det}(A) = 1(36 - 56) - 3(9 - 24) + (-5)(-7 + 12) \][/tex]
[tex]\[ \text{det}(A) = 1(-20) - 3(-15) + (-5)(5) \][/tex]
[tex]\[ \text{det}(A) = -20 + 45 - 25 \][/tex]
[tex]\[ \text{det}(A) = 0 \][/tex]
3. Determine the Nature of Solutions:
For a system of homogeneous linear equations, if the determinant of the coefficient matrix is zero, the system has either no solution or infinitely many solutions. Because our system is homogeneous, it always has the trivial solution (where \(n_1 = n_2 = n_3 = 0\)).
Since the determinant is zero, the system has a non-trivial solution (other than the all-zero solution). Therefore, the given system of equations has a non-trivial solution.
Conclusion:
The determinant of the coefficient matrix is 0.0. Thus, the given homogeneous system of equations has a non-trivial solution.
1. Form the Coefficient Matrix:
The coefficient matrix \(A\) for the given system of equations can be written as:
[tex]\[ A = \begin{bmatrix} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9 \end{bmatrix} \][/tex]
2. Calculate the Determinant of the Matrix \(A\):
The determinant of a 3x3 matrix:
[tex]\[ \text{det}(A) = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \][/tex]
is calculated using the formula:
[tex]\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]
For our matrix \(A\):
[tex]\[ A = \begin{bmatrix} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9 \end{bmatrix} \][/tex]
Applying the formula, the determinant \(\text{det}(A)\) is:
[tex]\[ \text{det}(A) = 1(4 \cdot 9 - (-8) \cdot (-7)) - 3(1 \cdot 9 - (-8) \cdot (-3)) + (-5)(1 \cdot (-7) - 4 \cdot (-3)) \][/tex]
Simplifying the expression, we get:
[tex]\[ \text{det}(A) = 1(36 - 56) - 3(9 - 24) + (-5)(-7 + 12) \][/tex]
[tex]\[ \text{det}(A) = 1(-20) - 3(-15) + (-5)(5) \][/tex]
[tex]\[ \text{det}(A) = -20 + 45 - 25 \][/tex]
[tex]\[ \text{det}(A) = 0 \][/tex]
3. Determine the Nature of Solutions:
For a system of homogeneous linear equations, if the determinant of the coefficient matrix is zero, the system has either no solution or infinitely many solutions. Because our system is homogeneous, it always has the trivial solution (where \(n_1 = n_2 = n_3 = 0\)).
Since the determinant is zero, the system has a non-trivial solution (other than the all-zero solution). Therefore, the given system of equations has a non-trivial solution.
Conclusion:
The determinant of the coefficient matrix is 0.0. Thus, the given homogeneous system of equations has a non-trivial solution.
