Question

Consider the system of linear equations
\[ x+y+z=1, \]\[ 2x+y+3z=6, \]\[ 3x+2y+kz=k+1. \]If the above system has no solution, then the value of \(k\) equals _______ (answer in integer).

Hint:

Eliminate \(x\) and \(y\) to get an equation of the form \((k-4)z=k-6\); the system has no solution when this reduces to \(0=\text{nonzero}\), that is at \(k=4\).

Answer 1

Correct Answer: 4

Step 1: Write down the three equations.
\[ x+y+z=1 \quad \text{...(i)} \]\[ 2x+y+3z=6 \quad \text{...(ii)} \]\[ 3x+2y+kz=k+1 \quad \text{...(iii)} \]

Step 2: Eliminate \(y\) using equations (i) and (ii).
From equation (i), \(y=1-x-z.\) Substitute into equation (ii): \[ 2x+(1-x-z)+3z=6 \]\[ x+2z+1=6 \]\[ x+2z=5. \quad \text{...(iv)} \] So \(x=5-2z.\)

Step 3: Express \(y\) in terms of \(z\).
\[ y=1-x-z=1-(5-2z)-z=1-5+2z-z=z-4. \]

Step 4: Substitute \(x\) and \(y\) into equation (iii).
\[ 3x+2y+kz=k+1 \]\[ 3(5-2z)+2(z-4)+kz=k+1 \]\[ 15-6z+2z-8+kz=k+1 \]\[ 7-4z+kz=k+1. \]

Step 5: Collect the \(z\) terms.
\[ z(k-4)=k+1-7 \]\[ z(k-4)=k-6. \quad \text{...(v)} \]

Step 6: Analyze when the system has no solution.
Equation (v) determines \(z\) uniquely whenever \(k\neq4\), since then \(z=\dfrac{k-6}{k-4},\) and \(x\), \(y\) follow uniquely from Steps 2 and 3. So for \(k\neq4\) the system has a unique solution, not no solution. When \(k=4\), equation (v) becomes \[ 0\cdot z=4-6=-2, \] that is, \(0=-2\), which is never true for any value of \(z\). This means the three equations are inconsistent, so the system has no solution.

Step 7: Final conclusion.
The system has no solution only when \(k=4.\)\[ \boxed{4} \]