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

Step 1: Understanding the Concept:
A system of equations $AX = B$ does not have a unique solution if the determinant of the coefficient matrix $A$ is zero ($|A| = 0$).

Step 2: Detailed Explanation:
1. Calculate the determinant $|A|$:
\[ |A| = \begin{vmatrix} \alpha & 2 & 3 \\ 2 & 3 & -\alpha \\ 3 & 5 & \alpha+1 \end{vmatrix} = 0 \]
2. Expand along the first row:
\[ \alpha[3(\alpha+1) - 5(-\alpha)] - 2[2(\alpha+1) - 3(-\alpha)] + 3[2(5) - 3(3)] = 0 \]
\[ \alpha[3\alpha + 3 + 5\alpha] - 2[2\alpha + 2 + 3\alpha] + 3[10 - 9] = 0 \]
\[ \alpha[8\alpha + 3] - 2[5\alpha + 2] + 3[1] = 0 \]
\[ 8\alpha^2 + 3\alpha - 10\alpha - 4 + 3 = 0 \]
\[ 8\alpha^2 - 7\alpha - 1 = 0 \]
3. Solve the quadratic equation:
\[ 8\alpha^2 - 8\alpha + \alpha - 1 = 0 \]
\[ 8\alpha(\alpha - 1) + 1(\alpha - 1) = 0 \]
\[ (8\alpha + 1)(\alpha - 1) = 0 \]
The roots are $\alpha = 1$ or $\alpha = -1/8$.
Since $\alpha$ must be an integer, $\alpha = 1$.

Step 3: Final Answer:
The value of $\alpha$ is 1.