Question

Consider the following system of equations:
$x - 2y + 3z = -1$
$x - 3y + 4z = 1$
$-2x + 4y - 6z = k$
The value of $k$ for which the system has infinitely many solutions is _______

• A. 0
• B. 1
• C. 2
• D. -1

Hint:

Notice that the third equation's left-hand side, $-2x + 4y - 6z$, is exactly $-2$ times the left-hand side of the first equation, $x - 2y + 3z$.
For the system to be consistent and have infinitely many solutions, the right-hand side of the third equation must also be $-2$ times the right-hand side of the first equation:
\[ k = -2 \times (-1) \implies k = 2 \]
This simple observation avoids the need for full row reduction.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks for the value of the parameter $k$ such that the given system of three linear equations with three variables has an infinite number of solutions.

Step 2: Key Formula or Approach:
According to the Rouché-Capelli theorem, a non-homogeneous system of linear equations $A X = B$ 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 rank is strictly less than the number of variables $n$:
\[ \text{rank}(A) = \text{rank}([A|B]) < n \]

Step 3: Detailed Explanation:

• Let us write the augmented matrix $[A|B]$ for the given system:
\[ [A|B] = \begin{pmatrix} 1 & -2 & 3 & | & -1 \\ 1 & -3 & 4 & | & 1 \\ -2 & 4 & -6 & | & k \end{pmatrix} \]

• We apply elementary row operations to reduce this augmented matrix to row echelon form.

• First, perform the row operation $R_2 \to R_2 - R_1$:
\[ \begin{pmatrix} 1 & -2 & 3 & | & -1 \\ 0 & -1 & 1 & | & 2 \\ -2 & 4 & -6 & | & k \end{pmatrix} \]

• Next, perform the row operation $R_3 \to R_3 + 2R_1$:
\[ \begin{pmatrix} 1 & -2 & 3 & | & -1 \\ 0 & -1 & 1 & | & 2\\ 0 & 0 & 0 & | & k - 2 \end{pmatrix} \]

• Now, we analyze the resulting echelon form.

• The third row of the coefficient matrix $A$ (the first three columns) contains only zeros, which means the rank of the coefficient matrix is:
\[ \text{rank}(A) = 2 \]

• For the system to be consistent (meaning at least one solution exists), the rank of the augmented matrix $[A|B]$ must also be equal to 2.

• If the last entry in the third row ($k - 2$) is non-zero, the rank of $[A|B]$ would be 3, making the system inconsistent (zero solutions).

• Therefore, we must have:
\[ k - 2 = 0 \implies k = 2 \]

• When $k = 2$, we have $\text{rank}(A) = \text{rank}([A|B]) = 2$, which is strictly less than the number of variables ($n = 3$).

• This condition yields a free variable, resulting in infinitely many solutions.

Step 4: Final Answer
Thus, the system has infinitely many solutions when $k = 2$, which corresponds to option (C).