Question

Let R and R³ denote the set of real numbers and the three dimensional vector space over it, respectively. The value of a for which the set of vectors
{[2-3 α], [3 -1 3], [1 −5 7]} 
does not form a basis of R³ 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

Correct Answer: 5

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.

Step 2: Form the matrix.
Arrange the given vectors as rows (or columns) of a matrix: \[ A = \begin{bmatrix} 2 & -3 & \alpha \\ 3 & -1 & 3 \\ 1 & -5 & 7 \end{bmatrix} \]

Step 3: Compute the determinant.
The determinant of \( A \) is: \[ \begin{vmatrix} 2 & -3 & \alpha \\ 3 & -1 & 3 \\ 1 & -5 & 7 \end{vmatrix} \] Expanding along the first row: \[ = 2 \left( (-1)(7) - (3)(-5) \right) - (-3) \left( (3)(7) - (3)(1) \right) + \alpha \left( (3)(-5) - (-1)(1) \right) \] Simplifying: \[ = 2(-7 + 15) + 3(21 - 3) + \alpha(-15 + 1) \] \[ = 2(8) + 3(18) - 14\alpha \] \[ = 16 + 54 - 14\alpha \] \[ = 70 - 14\alpha \]

Step 4: Determine when the vectors are dependent.
The determinant equals zero when the vectors are linearly dependent: \[ 70 - 14\alpha = 0 \] Solving for \( \alpha \): \[ \alpha = 5 \]

The vectors fail to form a basis of \( \mathbb{R}^3 \) when \( \alpha = 5 \).

Final Answer: \[ \boxed{5} \]