Question

Consider the following system of linear equations: 
\[ \begin{cases} x + y + 5z = 3, \\ x + 2y + mz = 5, \\ x + 2y + 4z = k. \end{cases} \] 

The system is consistent if 
 

• A. \( m \neq 4 \)
• B. \( k \neq 5 \)
• C. \( m = 4 \)
• D. \( k = 5 \)

Hint:

When analyzing system consistency, use Gaussian elimination and check when a zero row yields a contradiction like \( 0 = c \).

Answer 1

The Correct Option is A, D

Step 1: Finding consistency condition:

Write the augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 1 & 5 & 3 \\ 1 & 2 & m & 5 \\ 1 & 2 & 4 & k \end{array}\right]$$

Perform row operations:

$R_2 - R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 1 & 2 & 4 & k \end{array}\right]$$

$R_3 - R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 1 & -1 & k-3 \end{array}\right]$$

$R_3 - R_2$: $$\left[\begin{array}{ccc|c} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 0 & -1-(m-5) & k-3-2 \end{array}\right]$$

 $$= \left[\begin{array}{ccc|c} 1 & 1 & 5 & 3 \\ 0 & 1 & m-5 & 2 \\ 0 & 0 & 4-m & k-5 \end{array}\right]$$

Step 2: Consistency analysis:

For the system to be consistent, we need to avoid the situation where $0 = \text{non-zero}$ in the last row.

Case 1: If $4 - m \neq 0$ (i.e., $m \neq 4$):

The third row gives: $(4-m)z = k-5$

This has a solution for any value of $k$. The system is consistent for all $k$.

Case 2: If $4 - m = 0$ (i.e., $m = 4$):

The third row becomes: $0 = k - 5$

This is consistent only if $k = 5$.

Step 3: Conclusion:

The system is consistent if:

• $m \neq 4$ (for any value of $k$), OR
• $m = 4$ AND $k = 5$

Step 4: Evaluating options:

(A) $m \neq 4$: System is consistent for all $k$. TRUE 

(B) $k \neq 5$: Not sufficient alone; depends on $m$. FALSE 

(C) $m = 4$: Only consistent if $k = 5$. NOT ALWAYS TRUE 

(D) $k = 5$: System is consistent for all $m$ (including $m = 4$). TRUE 

Answer: (A) and (D) are correct