Question

Let \(\mathbb{R}\) and \(\mathbb{R}^3\) denote the set of real numbers and the three-dimensional vector space over it, respectively. The value of \(\alpha\) for which the set of vectors: \[ \{[2 \,\, -3 \,\, \alpha], \, [3 \,\, -1 \,\, 3], \, [1 \,\, -5 \,\, 7]\} \] does not form a basis of \(\mathbb{R}^3\) is \(\_\_\_\_\).

Hint:

To verify if vectors form a basis in \( \mathbb{R}^3 \), calculate the determinant of the associated matrix. A zero determinant indicates linear dependence.

Answer 1

Step 1: Basis condition in \( \mathbb{R}^3 \).
A set of vectors forms a basis for \( \mathbb{R}^3 \) if the vectors are linearly independent. This condition is satisfied if the determinant of the matrix formed by these vectors is non-zero. \medskip Step 2: Form the matrix.
Arrange the given vectors as rows (or columns) of a matrix: \[ A = \begin{bmatrix} 2 & -3 & \alpha
3 & -1 & 3
1 & -5 & 7 \end{bmatrix}. \] \medskip Step 3: Compute the determinant.
The determinant of \(A\) is: \[ \left| \begin{matrix} 2 & -3 & \alpha
3 & -1 & 3
1 & -5 & 7 \end{matrix} \right|. \] Expanding along the first row: \[ = 2 \left( (-1)(7) - (3)(-5) \right) - (-3) \left( (3)(7) - (3)(1) \right) + \alpha \left( (3)(-5) - (-1)(1) \right). \] Simplify the terms: \[ = 2(-7 + 15) + 3(21 - 3) + \alpha(-15 + 1), \] \[ = 2(8) + 3(18) - 14\alpha, \] \[ = 16 + 54 - 14\alpha, \] \[ = 70 - 14\alpha. \] \medskip Step 4: Determine when the vectors are dependent.
The determinant equals zero when the vectors are linearly dependent: \[ 70 - 14\alpha = 0. \] Solving for \(\alpha\): \[ \alpha = 5. \] \medskip The vectors fail to form a basis of \( \mathbb{R}^3 \) when \( \alpha = 5 \). Final Answer: \[ \boxed{5} \]