Key Concept: Conditions for Infinitely Many Solutions for a System of Linear Equations

For a system of linear equations in three variables: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$

This system can be represented in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable vector, and $B$ is the constant vector. $A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$ For such a system to have infinitely many solutions, two crucial conditions must be satisfied:

1. The determinant of the coefficient matrix $A$ must be zero. That is, $\text{det}(A) = 0$.

   • Why? If $\text{det}(A) \neq 0$, then $A^{-1}$ exists, and the system would have a unique solution given by $X = A^{-1}B$. For infinitely many solutions, the matrix $A$ must be singular (non-invertible), which means its determinant must be zero.

2. The system must be consistent. When $\text{det}(A) = 0$, the system could either have infinitely many solutions or no solution. To distinguish between these two cases, we use Cramer's Rule. For infinitely many solutions, all the determinants $D_x, D_y, D_z$ (formed by replacing a column of $A$ with the constant vector $B$) must also be zero.

   • Why? If $\text{det}(A) = 0$ but at least one of $D_x, D_y, D_z$ is non-zero, the system would be inconsistent, meaning it has no solution. For infinitely many solutions, the system must be consistent, leading to the condition $D_x=D_y=D_z=0$.

In summary, for a system of $n$ linear equations in $n$ variables to have infinitely many solutions, the following conditions must hold: $\text{det}(A) = 0 \quad \text{AND} \quad D_x = 0 \quad \text{AND} \quad D_y = 0 \quad \text{AND} \quad D_z = 0$

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Step 1: Formulate the Coefficient Matrix and Constant Vector

The given system of linear equations is:

1. $3x - y + 4z = 3$
2. $x + 2y - 3z = -2$
3. $6x + 5y + kz = -3$

From these equations, we can identify the coefficient matrix $A$ and the constant vector $B$: $A = \begin{pmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ -2 \\ -3 \end{pmatrix}$

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Step 2: Apply the Condition $\text{det}(A)=0$ to find candidate values for $k$

For the system to have infinitely many solutions, the determinant of the coefficient matrix $A$ must be zero. We calculate $\text{det}(A)$ by expanding along the first row: $\text{det}(A) = \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix}$ $\text{det}(A) = 3 \begin{vmatrix} 2 & -3 \\ 5 & k \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3 \\ 6 & k \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 6 & 5 \end{vmatrix}$

• Why? We expand the determinant using cofactor expansion to express $\text{det}(A)$ in terms of $k$.

Now, let's evaluate the $2 \times 2$ determinants: $\text{det}(A) = 3((2)(k) - (-3)(5)) + 1((1)(k) - (-3)(6)) + 4((1)(5) - (2)(6))$ $\text{det}(A) = 3(2k + 15) + 1(k + 18) + 4(5 - 12)$ $\text{det}(A) = 6k + 45 + k + 18 + 4(-7)$ $\text{det}(A) = 7k + 63 - 28$ $\text{det}(A) = 7k + 35$

• Why? We simplify the expression to find a linear equation in $k$.

Now, we set $\text{det}(A) = 0$ to find the value(s) of $k$ that satisfy this necessary condition: $7k + 35 = 0$ $7k = -35$ $k = -5$

• Why? This step provides the candidate value for $k$. This value makes the coefficient matrix singular, which is essential for either infinitely many solutions or no solution.

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Step 3: Verify Consistency by checking $D_x=D_y=D_z=0$ for $k=-5$

We found $k=-5$ as a potential value. To confirm that for this value of $k$ the system has infinitely many solutions (and not no solution), we must verify that the system is consistent by checking if $D_x=0$, $D_y=0$, and $D_z=0$ when $k=-5$.

Let's form the matrices for $D_x, D_y, D_z$ by replacing the respective columns of $A$ with the constant vector $B$, with $k=-5$: $D_x = \begin{vmatrix} 3 & -1 & 4 \\ -2 & 2 & -3 \\ -3 & 5 & -5 \end{vmatrix}$ $D_y = \begin{vmatrix} 3 & 3 & 4 \\ 1 & -2 & -3 \\ 6 & -3 & -5 \end{vmatrix}$ $D_z = \begin{vmatrix} 3 & -1 & 3 \\ 1 & 2 & -2 \\ 6 & 5 & -3 \end{vmatrix}$

• Why? These determinants are crucial for applying Cramer's Rule. If $\text{det}(A)=0$ and any of these ($D_x, D_y, D_z$) are non-zero, the system would be inconsistent (no solution).

Now, let's calculate each determinant:

Calculate $D_x$: $D_x = 3 \begin{vmatrix} 2 & -3 \\ 5 & -5 \end{vmatrix} - (-1) \begin{vmatrix} -2 & -3 \\ -3 & -5 \end{vmatrix} + 4 \begin{vmatrix} -2 & 2 \\ -3 & 5 \end{vmatrix}$ $D_x = 3((2)(-5) - (-3)(5)) + 1((-2)(-5) - (-3)(-3)) + 4((-2)(5) - (2)(-3))$ $D_x = 3(-10 + 15) + 1(10 - 9) + 4(-10 + 6)$ $D_x = 3(5) + 1(1) + 4(-4)$ $D_x = 15 + 1 - 16$ $D_x = 0$

Calculate $D_y$: $D_y = 3 \begin{vmatrix} -2 & -3 \\ -3 & -5 \end{vmatrix} - 3 \begin{vmatrix} 1 & -3 \\ 6 & -5 \end{vmatrix} + 4 \begin{vmatrix} 1 & -2 \\ 6 & -3 \end{vmatrix}$ $D_y = 3((-2)(-5) - (-3)(-3)) - 3((1)(-5) - (-3)(6)) + 4((1)(-3) - (-2)(6))$ $D_y = 3(10 - 9) - 3(-5 + 18) + 4(-3 + 12)$ $D_y = 3(1) - 3(13) + 4(9)$ $D_y = 3 - 39 + 36$ $D_y = 0$

Calculate $D_z$: $D_z = 3 \begin{vmatrix} 2 & -2 \\ 5 & -3 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -2 \\ 6 & -3 \end{vmatrix} + 3 \begin{vmatrix} 1 & 2 \\ 6 & 5 \end{vmatrix}$ $D_z = 3((2)(-3) - (-2)(5)) + 1((1)(-3) - (-2)(6)) + 3((1)(5) - (2)(6))$ $D_z = 3(-6 + 10) + 1(-3 + 12) + 3(5 - 12)$ $D_z = 3(4) + 1(9) + 3(-7)$ $D_z = 12 + 9 - 21$ $D_z = 0$

Since $\text{det}(A)=0$, $D_x=0$, $D_y=0$, and $D_z=0$ when $k=-5$, all conditions for infinitely many solutions are met.

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Step 4: State the Final Answer

The value of $k$ for which the given system of linear equations has infinitely many solutions is $k=-5$.

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Tips and Common Mistakes

• Do Not Stop at $\text{det}(A)=0$: A common error is to only calculate $\text{det}(A)=0$ and assume that the resulting value of $k$ will guarantee infinitely many solutions. This is a necessary condition but not sufficient. $\text{det}(A)=0$ can also lead to a system having no solution (inconsistent system). Always verify the consistency condition by checking that $D_x=D_y=D_z=0$.
• Alternative Method (Rank Method): For a system $AX=B$ of $n$ linear equations in $n$ variables, it has infinitely many solutions if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A|B]$, and this common rank is less than $n$. In this problem, $n=3$, so we would need $\text{rank}(A) = \text{rank}([A|B]) < 3$. This method involves row-reducing the augmented matrix. While powerful, for 3x3 systems, using determinants (Cramer's Rule conditions) can often be more straightforward if calculations are done carefully.
• Calculation Errors: Determinant calculations involve multiple multiplications and subtractions, making them prone to arithmetic and sign errors. Double-check each step, especially when expanding $2 \times 2$ and $3 \times 3$ determinants.

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Summary and Key Takeaway

To solve for a parameter that results in a system of linear equations having infinitely many solutions, we must apply two fundamental conditions:

1. The determinant of the coefficient matrix ($\text{det}(A)$) must be zero. This condition identifies the values of the parameter for which the system is not uniquely solvable.
2. For the identified parameter values, the system must also be consistent. This is verified by checking that all related determinants ($D_x, D_y, D_z$ in Cramer's Rule) are also zero. By systematically applying these conditions and performing careful calculations, we found that $k=-5$ is the specific value that satisfies both criteria, ensuring the given system has infinitely many solutions.

The final answer is $\boxed{-5}$.