Question

If the system of equations αx + y + z = 5, x + 2y + 3z = 4, x + 3y +5z = β has infinitely many solutions, then the ordered pair (α, β) is equal to:

• A. (1, –3)
• B. (–1, 3)
• C. (1, 3)
• D. (–1, –3)

Answer 1

The Correct Option is C

 To solve this problem, we need to determine the values of \(\alpha\) and \(\beta\) such that the given system of equations has infinitely many solutions. This requires examining the conditions under which a system of equations is consistent and dependent.

The given system is:

1. \(\alpha x + y + z = 5\)
2. \(x + 2y + 3z = 4\)
3. \(x + 3y + 5z = \beta\)

For this system to have infinitely many solutions, the equations must be linearly dependent. The condition for linear dependence in a system of three equations with three unknowns is that the determinant of the coefficient matrix must be zero. However, since \(\alpha\) and \(\beta\) are not variables of equation coefficients but constant terms, we will instead focus on ensuring all equations are scalar multiples of each other or obtain a contradiction that \(\text{rank}(A) = \text{rank}([A|b]) < 3\), where \(A\) is the coefficient matrix and \(b\) is the constants.

Consider the second row (equation 2) and third row (equation 3). For these to be dependent with the first:

• The coefficient ratios between rows must be equal, so comparing rows 2 and 3: coefficients of \(x\), \(y\), and \(z\) are \(1, 2, 3\) and \(1, 3, 5\). These should match coefficients from row corresponding the coefficients \(\alpha, 1, 1\).

Set the proportion:

1. From equations 2 and 3, equate \(x\): \(1/1 = 1/1\) (Trivially equal)
2. Equate \(y\): \(2/1 = 3/1\).

Now relate constant terms for verification:

• The consistent system must also satisfy \(4/\beta = 5/5\). Therefore, \(\beta = 3\).

To find \(\alpha\), equate first and second equation with discovered independent constant:

• \(\alpha = 1\)

Substitute back: system infinite if

• \(\alpha = 1\) and \(\beta = 3\).

Therefore, the ordered pair that satisfies having infinitely many solutions is (1, 3).