Key Concept: Homogeneous System of Linear Equations

A system of linear equations is called homogeneous if all the constant terms on the right-hand side are zero. Such a system can be represented in matrix form as $AX = 0$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $0$ is the column vector of zeros.

For a homogeneous system $AX = 0$ with a square coefficient matrix $A$:

1. Always Consistent: A homogeneous system is always consistent, meaning it always has at least one solution. The most obvious solution is the trivial solution, where all variables are zero ($x=0, y=0, z=0$). Therefore, the option "no solution" is never possible for a homogeneous system.
2. Determinant Test:
   • If the determinant of the coefficient matrix, $\det(A)$, is non-zero ($\det(A) \neq 0$), the system has only the trivial solution ($x=0, y=0, z=0$).
   • If the determinant of the coefficient matrix, $\det(A)$, is zero ($\det(A) = 0$), the system has infinitely many solutions. These solutions include the trivial solution and also non-trivial solutions (where at least one variable is non-zero).

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Step 1: Form the Coefficient Matrix and Calculate its Determinant

The given system of linear equations is:

1. $7x + 6y - 2z = 0$
2. $3x + 4y + 2z = 0$
3. $x - 2y - 6z = 0$

Why this step? To determine the nature of solutions (trivial or infinitely many), we first need to identify the coefficient matrix and calculate its determinant. The determinant value is the primary criterion for homogeneous systems.

This is a homogeneous system because all the constant terms on the right-hand side are zero. We form the coefficient matrix $A$ from the coefficients of $x, y, z$: $A = \begin{pmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{pmatrix}$

Now, we calculate the determinant of this matrix, $\Delta = \det(A)$. We will use the cofactor expansion along the first row for clarity. $\Delta = \begin{vmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{vmatrix}$ Expanding along the first row: $\Delta = 7 \begin{vmatrix} 4 & 2 \\ -2 & -6 \end{vmatrix} - 6 \begin{vmatrix} 3 & 2 \\ 1 & -6 \end{vmatrix} + (-2) \begin{vmatrix} 3 & 4 \\ 1 & -2 \end{vmatrix}$ Calculating the $2 \times 2$ determinants: $\Delta = 7((4)(-6) - (2)(-2)) - 6((3)(-6) - (2)(1)) - 2((3)(-2) - (4)(1))$ $\Delta = 7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4)$ $\Delta = 7(-20) - 6(-20) - 2(-10)$ $\Delta = -140 + 120 + 20$ $\Delta = 0$

Step 2: Determine the Nature of Solutions

Why this step? The value of the determinant dictates whether the homogeneous system has a unique trivial solution or infinitely many solutions.

Since $\det(A) = 0$, the homogeneous system of linear equations has infinitely many solutions. These solutions include the trivial solution ($x=0, y=0, z=0$) and also non-trivial solutions. This conclusion immediately rules out option (D) "only the trivial solution". As discussed in the Key Concept, "no solution" (option A) is impossible for a homogeneous system.

Step 3: Find the Relationships Between Variables for Non-Trivial Solutions

Why this step? Since there are infinitely many solutions, we need to find the specific relationships between $x, y,$ and $z$ that define these solutions, in order to match them with the given options (B) or (C).

Since $\det(A) = 0$, the equations are not linearly independent. We can express the variables in terms of one another. We can use any two linearly independent equations to find the relationships. Let's use the first two equations:

1. $7x + 6y - 2z = 0$
2. $3x + 4y + 2z = 0$

Add equation (1) and equation (2) to eliminate $z$: $(7x + 6y - 2z) + (3x + 4y + 2z) = 0 + 0$ $10x + 10y = 0$ $10(x+y) = 0$ $x + y = 0 \implies \boxed{x = -y}$

Now substitute $x = -y$ into equation (2): $3(-y) + 4y + 2z = 0$ $-3y + 4y + 2z = 0$ $y + 2z = 0 \implies \boxed{y = -2z}$

Now we have $x$ and $y$ in terms of $z$: From $y = -2z$, substitute this into $x = -y$: $x = -(-2z)$ $x = 2z$

So, the solutions $(x, y, z)$ satisfy the relationships: $x = 2z$ $y = -2z$ $z = z$ (where $z$ can be any real number)

We can express the general solution as $(2k, -2k, k)$ for any real number $k$.

Step 4: Match with Options

Why this step? We compare our derived relationships with the given options to find the correct one.

Our derived relationships are $x = 2z$ and $y = -2z$. Let's check the options for infinitely many solutions:

• (B) infinitely many solutions, (x, y, z) satisfying $y = 2z$. This does not match our derived $y = -2z$. So, (B) is incorrect.
• (C) infinitely many solutions, (x, y, z) satisfying $x = 2z$. This perfectly matches our derived relationship $x = 2z$.

Therefore, the system has infinitely many solutions, and these solutions satisfy $x = 2z$.

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Summary and Key Takeaway

For a homogeneous system of linear equations ($AX=0$):

1. It is always consistent, meaning it always has at least the trivial solution ($x=0, y=0, z=0$). Thus, "no solution" is never possible.
2. Calculate the determinant of the coefficient matrix, $\det(A)$.
   • If $\det(A) \neq 0$, there is a unique trivial solution.
   • If $\det(A) = 0$, there are infinitely many solutions (including non-trivial ones).
3. If $\det(A) = 0$, you must further solve the system (using elimination or row operations) to find the specific relationships between variables that characterize these infinite solutions, and then match them with the given options.

In this problem, $\det(A) = 0$, indicating infinitely many solutions. By solving the system, we found that these solutions satisfy $x = 2z$ (and $y = -2z$). Thus, option (C) is the correct description of the solution set.