Question

For $\alpha, \beta \in R$, suppose the system of linear equations $ x-y+z=5 $ $2 x+2 y+\alpha z=8 $ $3 x-y+4 z=\beta$ has infinitely many solutions Then $\alpha$ and $\beta$ are the roots of

• A. $x^2-18 x+56=0$
• B. $x^2+14 x+24=0$
• C. $x^2-10 x+16=0$
• D. $x^2+18 x+56=0$

Answer 1

The Correct Option is A

To solve this problem, we need to ensure that the given system of linear equations has infinitely many solutions. A system of linear equations has infinitely many solutions if the equations are consistent and dependent. Let's analyze the given system of equations:

• Equation 1: \(x - y + z = 5\) 
• Equation 2: \(2x + 2y + \alpha z = 8\)
• Equation 3: \(3x - y + 4z = \beta\)

First, let's express these equations in matrix form:

\(\begin{bmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{bmatrix}\)

\(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\)

=

\(\begin{bmatrix} 5 \\ 8 \\ \beta \end{bmatrix}\)

To have infinitely many solutions, the rank of the coefficient matrix should be equal to the rank of the augmented matrix and less than the number of variables.

Let's first row-reduce the coefficient matrix:

1. Using Equation 1, eliminate \(x\) from Equations 2 and 3.

Subtract 2 times Equation 1 from Equation 2:

\((2x + 2y + \alpha z) - 2(x - y + z) = 8 - 2 \times 5\)

Simplifies to:

\(4y + (\alpha - 2)z = -2\)

Subtract 3 times Equation 1 from Equation 3:

\((3x - y + 4z) - 3(x - y + z) = \beta - 3 \times 5\)

Simplifies to:

\(2y + z = \beta - 15\)

The reduced system becomes:

• \(x - y + z = 5\)
• \(4y + (\alpha - 2)z = -2\)
• \(2y + z = \beta - 15\)

For infinitely many solutions, the second and third equations must be dependent. Thus, we compare the coefficients:

The coefficient ratio of \(y\) and constant term must be equal:

\(\frac{4}{2} = \frac{\alpha - 2}{1} = \frac{-2}{\beta - 15}\)

From \(\frac{4}{2} = \frac{\alpha - 2}{1}\), we solve:

\(2(\alpha - 2) = 4\), so \(\alpha = 4\).

From \(\frac{4}{2} = \frac{-2}{\beta - 15}\), we solve:

\(2(-2) = 4(\beta - 15)\), so \(-4 = 4\beta - 60\)

\(4\beta = 56\), so \(\beta = 14\).

Thus, \(\alpha = 4\) and \(\beta = 14\).

Now, we identify the correct quadratic equation whose roots are 4 and 14. The sum of the roots \(4 + 14 = 18\) and their product is \(4 \times 14 = 56\). The required quadratic equation is:

\(x^2 - 18x + 56 = 0\)

This matches the option \(x^2 - 18x + 56 = 0\), confirming it as the correct answer.