Question

A system of equations is given in the matrix form as $\begin{bmatrix} \alpha & 2 & 3 \\ 2 & 3 & -\alpha \\ 3 & 5 & \alpha+1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}$ where $\alpha$ is an integer. If the system of equations does not have a unique solution, then the value of $\alpha$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Hint:

For matrix determinant questions in competitive exams, if one root is a clear integer and the other is a fraction, the integer is almost always the intended answer.

Answer 1

The Correct Option is A

To determine when the given system of equations does not have a unique solution, we must examine the condition for the uniqueness of the solution in a system of linear equations.

The given system is represented in matrix form as:

\(\begin{bmatrix} \alpha & 2 & 3 \\ 2 & 3 & -\alpha \\ 3 & 5 & \alpha + 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}\)

A system of equations has a unique solution if the determinant of the coefficient matrix is non-zero.

Let's calculate the determinant of the matrix:

\(\text{det}(A) = \begin{vmatrix} \alpha & 2 & 3 \\ 2 & 3 & -\alpha \\ 3 & 5 & \alpha + 1 \end{vmatrix}\)

Using the rule for calculating the determinant of a 3x3 matrix,

\(\text{det}(A) = \alpha [(3)(\alpha + 1) - (-\alpha)(5)] - 2 [2(\alpha + 1) - (-\alpha)(3)] + 3 [2(5) - 3(3)]\)

Simplify each term:

• \(\alpha [3\alpha + 3 + 5\alpha] = \alpha [8\alpha + 3]\)
• \(-2 [2\alpha + 2 + 3\alpha] = -2 [5\alpha + 2]\)
• \(3 [10 - 9] = 3 [1] = 3\)

Plug these back into the determinant expression:

\(\alpha (8\alpha + 3) - 2 (5\alpha + 2) + 3\)

Break down further:

• \(8\alpha^2 + 3\alpha\)
• \(-10\alpha - 4\)
• \(+3\)

Combine like terms:

\(8\alpha^2 + 3\alpha - 10\alpha - 4 + 3 = 8\alpha^2 - 7\alpha - 1\)

For the system to not have a unique solution, this determinant must be zero:

\(8\alpha^2 - 7\alpha - 1 = 0\)

Solving this quadratic equation using the quadratic formula, \(\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = -7\), \(c = -1\):

\(\alpha = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 8 \times (-1)}}{2 \times 8}\)

Simplify:

\(\alpha = \frac{7 \pm \sqrt{49 + 32}}{16}\)

\(\alpha = \frac{7 \pm \sqrt{81}}{16}\)

\(\alpha = \frac{7 \pm 9}{16}\)

This results in two potential solutions for \(\alpha\):

• \(\alpha = \frac{16}{16} = 1\)
• \(\alpha = \frac{-2}{16} = -\frac{1}{8}\)

Since \(\alpha\) needs to be an integer, \(\alpha = 1\) is the only valid solution.

Therefore, the value of \(\alpha\) that causes the system to not have a unique solution is 1.