Question

Let the system of equations be αu+w=0, u+αν =0, v+αw=0, where a ∈ ℜ. Then the system has infinite solutions if a =_____ (in integer).

Answer 1

Correct Answer: -1

The system of linear equations has infinite solutions if and only if the determinant of the coefficient matrix is equal to zero ($\det(\mathbf{A})=0$) and the system is homogeneous (which it is, since all equations equal zero).

$\text{1. Formulate the Coefficient Matrix}$

The given system of equations is:

$$\begin{aligned} \alpha u + 0v + 1w &= 0 \\ 1u + \alpha v + 0w &= 0 \\ 0u + 1v + \alpha w &= 0 \end{aligned}$$

The coefficient matrix $\mathbf{A}$ is:

$$\mathbf{A} = \begin{pmatrix} \alpha & 0 & 1 \\ 1 & \alpha & 0 \\ 0 & 1 & \alpha \end{pmatrix}$$

$\text{2. Calculate the Determinant}$

For the system to have infinite non-trivial solutions, the determinant of the coefficient matrix must be zero: $\det(\mathbf{A}) = 0$.

We calculate $\det(\mathbf{A})$ using the cofactor expansion along the first row:

$$\det(\mathbf{A}) = \alpha \begin{vmatrix} \alpha & 0 \\ 1 & \alpha \end{vmatrix} - 0 \begin{vmatrix} 1 & 0 \\ 0 & \alpha \end{vmatrix} + 1 \begin{vmatrix} 1 & \alpha \\ 0 & 1 \end{vmatrix}$$

$$\det(\mathbf{A}) = \alpha [(\alpha)(\alpha) - (0)(1)] - 0 + 1 [(1)(1) - (\alpha)(0)]$$

$$\det(\mathbf{A}) = \alpha (\alpha^2) + 1 (1)$$

$$\det(\mathbf{A}) = \alpha^3 + 1$$

$\text{3. Solve for } \alpha$

Set the determinant equal to zero to find the values of $\alpha$ that lead to infinite solutions:

$$\det(\mathbf{A}) = \alpha^3 + 1 = 0$$

$$\alpha^3 = -1$$

The solutions for $\alpha$ are the cube roots of $-1$.

$$\alpha = \sqrt[3]{-1}$$

The cube roots of $-1$ are:

Real root: $\alpha = -1$

Complex roots: $\alpha = \frac{1 + i\sqrt{3}}{2}$ and $\alpha = \frac{1 - i\sqrt{3}}{2}$

Since the question asks for the value of $\alpha$ as an integer ($\alpha \in \mathbb{R}$ is given, but the answer must be an integer), we choose the real, integer root.

$$\alpha = -1$$

$$\text{The system has infinite solutions if } \alpha = \mathbf{-1}$$