Question

The system of linear equations \[ \begin{cases} x + y + z = 6 \\ 2x + 5y + az = 36 \\ x + 2y + 3z = b \end{cases} \] has:

• A. Infinitely many solutions for \( a = 8 \) and \( b = 16 \)
• B. Unique solution for \( a = 8 \) and \( b = 16 \)
• C. Unique solution for \( a = 8 \) and \( b = 14 \)
• D. Infinitely many solutions for \( a = 8 \) and \( b = 14 \)

Hint:

For linear systems, infinitely many solutions occur when equations are dependent but consistent.

Answer 1

The Correct Option is D

To determine the nature of solutions for the given system of linear equations, we will use the method of reduction and comparison. The system of equations is: 

\(\begin{cases} x + y + z = 6 \\ 2x + 5y + az = 36 \\ x + 2y + 3z = b \end{cases}\)

We need to find the values of \( a \) and \( b \) for which the system has infinitely many solutions.

1. First, let's write these equations in an augmented matrix form:

1	1	1	|	6
2	5	a	|	36
1	2	3	|	b

1. To have infinitely many solutions, the system should either have dependent equations or more than one consistent solutions. Here, we should look for values that make any two rows equivalent or similar.
2. Perform row operations to check the relations among the equations. From Row 1 (\(R_1\)) and Row 3 (\(R_3\)), subtract \(R_1\) from \(R_3\):

\((R_3 - R_1) \Rightarrow (x + 2y + 3z) - (x + y + z) = b - 6 \Rightarrow y + 2z = b - 6\)

1. Now, from Row 2 (\(R_2\)) and Row 1 (\(R_1\)), subtract \(2 \times R_1\) from \(R_2\):

\((R_2 - 2R_1) \Rightarrow (2x + 5y + az) - 2(x + y + z) = 36 - 2(6) \Rightarrow (5y + az - 2y - 2z) = 36 - 12\)

\(3y + (a-2)z = 24\)

1. For the system to have infinitely many solutions, these resultant equations should describe the same plane or lines should be parallel, i.e., there should be a proportional relationship. Let's equate the coefficients from steps 3 and 4 by expressing \(z\) in terms of \(y\) for each:

\(y + 2z = b - 6\ \Rightarrow z = \frac{b - 6 - y}{2}\)

\(3y + (a-2)z = 24\ \Rightarrow z = \frac{24 - 3y}{a-2}\)

Equating the two expressions for \(z\), we get:

\(\frac{b - 6 - y}{2} = \frac{24 - 3y}{a-2}\)

1. Cross-multiply and simplify to find conditions on \(a\) and \(b\):

\((b - 6 - y)(a-2) = 2(24 - 3y)\)

\((a-2)(b-6) - (a-2)y = 48 - 6y\)

Match coefficients for terms involving \(y\), and the constant terms:

\(-(a-2) = -6 \Rightarrow a-2 = 6 \Rightarrow a = 8\)

Match the constant terms:

\((a-2)(b-6) = 48 \Rightarrow 6(b-6) = 48 \Rightarrow b-6 = 8 \Rightarrow b = 14\)

This confirms that for \(a = 8\) and \(b = 14\), the system has infinitely many solutions.

Therefore, the correct answer is: Infinitely many solutions for \( a = 8 \) and \( b = 14 \).