1. Understanding the Conditions for Infinitely Many Solutions

For a system of linear equations in three variables, say $a_1x + b_1y + c_1z = d_1$, $a_2x + b_2y + c_2z = d_2$, and $a_3x + b_3y + c_3z = d_3$, represented in matrix form as $AX=B$, there are specific conditions for it to have infinitely many solutions.

Let the coefficient matrix be $A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$, and the constant matrix be $B = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$.

We define the following determinants:

• $\Delta = \det(A)$
• $\Delta_x = \det(A_x)$, where $A_x$ is formed by replacing the first column of $A$ with $B$.
• $\Delta_y = \det(A_y)$, where $A_y$ is formed by replacing the second column of $A$ with $B$.
• $\Delta_z = \det(A_z)$, where $A_z$ is formed by replacing the third column of $A$ with $B$.

A system of linear equations has infinitely many solutions if and only if:

1. $\Delta = 0$ (The determinant of the coefficient matrix is zero).
2. $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$ (All determinants formed by replacing a column of $A$ with $B$ are also zero).

Why these conditions?

• If $\Delta \neq 0$, the system has a unique solution (by Cramer's Rule).
• If $\Delta = 0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system is inconsistent and has no solution.
• The conditions $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$ imply that the equations are linearly dependent and consistent, meaning they represent planes that either coincide or intersect along a common line, leading to infinitely many solutions.

2. Representing the System in Matrix Form

The given system of linear equations is: $2x + y - z = 7$ $x - 3y + 2z = 1$ $x + 4y + \delta z = k$

We can write this system in the standard matrix form $AX=B$, where:

• $A$ is the coefficient matrix.
• $X$ is the column matrix of variables.
• $B$ is the column matrix of constants.

$A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 7 \\ 1 \\ k \end{pmatrix}$

3. Determining $\delta$ using $\Delta = 0$

For the system to have infinitely many solutions, the first condition is that the determinant of the coefficient matrix, $\Delta = \det(A)$, must be zero.

Let's calculate $\Delta$: $\Delta = \left| {\begin{matrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{matrix} } \right|$

We expand this $3 \times 3$ determinant along the first row for calculation: $\Delta = 2 \cdot ((-3)(\delta) - (2)(4)) - 1 \cdot ((1)(\delta) - (2)(1)) + (-1) \cdot ((1)(4) - (-3)(1))$ $\Delta = 2(-3\delta - 8) - 1(\delta - 2) - 1(4 + 3)$ $\Delta = -6\delta - 16 - \delta + 2 - 7$ $\Delta = -7\delta - 21$

Now, we apply the condition $\Delta = 0$: $-7\delta - 21 = 0$ $-7\delta = 21$ $\delta = \frac{21}{-7}$ $\boxed{\delta = -3}$

Explanation: By setting $\Delta = 0$, we ensure that the system does not have a unique solution. This is a prerequisite for infinitely many solutions (or no solution). This step determines the specific value of $\delta$ that makes the coefficient matrix singular.

4. Determining $k$ using $\Delta_x = 0$

Now that we have found $\delta = -3$, we need to satisfy the second set of conditions: $\Delta_x = 0$, $\Delta_y = 0$, and $\Delta_z = 0$. We only need to use one of these conditions to find $k$, as they must all hold true for infinite solutions. Let's use $\Delta_x = 0$.

First, we construct the matrix $A_x$ by replacing the first column of $A$ with the constant matrix $B$, and substitute $\delta = -3$: $A_x = \begin{pmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{pmatrix}$

Now, we calculate $\Delta_x = \det(A_x)$: $\Delta_x = \left| {\begin{matrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{matrix} } \right|$

Expanding along the first row: $\Delta_x = 7 \cdot ((-3)(-3) - (2)(4)) - 1 \cdot ((1)(-3) - (2)(k)) + (-1) \cdot ((1)(4) - (-3)(k))$ $\Delta_x = 7(9 - 8) - 1(-3 - 2k) - 1(4 + 3k)$ $\Delta_x = 7(1) - (-3 - 2k) - (4 + 3k)$ $\Delta_x = 7 + 3 + 2k - 4 - 3k$ $\Delta_x = 10 - 4 - k$ $\Delta_x = 6 - k$

Finally, we apply the condition $\Delta_x = 0$: $6 - k = 0$ $\boxed{k = 6}$

Explanation: Setting $\Delta_x = 0$ (along with $\Delta = 0$) ensures that the system is consistent. If $\Delta = 0$ and any of $\Delta_x, \Delta_y, \Delta_z$ were non-zero, the system would be inconsistent (no solutions). This step allows us to determine the value of $k$ that guarantees consistency when $\Delta=0$.

Common Mistake Alert: Some students might stop after finding $\Delta=0$ and assume that's enough. Remember, $\Delta=0$ alone could mean either "no solution" or "infinitely many solutions". The additional conditions ($\Delta_x = \Delta_y = \Delta_z = 0$) are crucial to distinguish between these two cases.

5. Calculating $\delta + k$

We found $\delta = -3$ and $k = 6$. Now, we calculate their sum: $\delta + k = -3 + 6$ $\delta + k = 3$

The final answer is $\boxed{3}$.

6. Summary and Key Takeaways

To solve systems of linear equations for conditions like "infinitely many solutions" or "no solution," Cramer's Rule and the properties of determinants are essential.

• For infinitely many solutions: Both $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$ must be satisfied.
• For no solution: $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
• For a unique solution: $\Delta \neq 0$.

In this problem, we systematically applied these conditions. First, $\Delta=0$ helped us find $\delta$. Then, one of the consistency conditions ($\Delta_x=0$) allowed us to determine $k$. Always double-check your determinant calculations, as a small error can propagate and lead to an incorrect answer.