Question

If \[ \{(2,-3,5),\ (1,\alpha,7),\ (3,\beta,3)\} \] is not a basis of the vector space \(\mathbb{R}^3\), then \((\alpha,\beta)\) lies on the locus:

• A. \(x^{2}+y=5\)
• B. \(y^{2}+x=4\)
• C. \(x+y+6=0\)
• D. \(x^{2}+y^{2}=16\)

Hint:

Whenever vectors are asked to form a basis of \(\mathbb R^3\), immediately construct a \(3\times3\) matrix and check its determinant. A zero determinant indicates linear dependence.

Answer 1

The Correct Option is C

Concept: A set of three vectors in \(\mathbb R^3\) forms a basis if and only if the vectors are linearly independent. For three vectors arranged as rows (or columns) of a matrix, linear independence is determined by the determinant. \[ \det(A)\neq 0 \] implies the vectors form a basis, while \[ \det(A)=0 \] implies the vectors are linearly dependent and hence do not form a basis. Therefore, we compute the determinant and impose the condition that it must be zero.

Step 1: Form the matrix using the given vectors.
\[ A= \begin{bmatrix} 2 & -3 & 5 \\ 1 & \alpha & 7\\ 3 & \beta & 3 \end{bmatrix} \] Since the vectors are not a basis, \[ |A|=0. \]

Step 2: Evaluate the determinant.
Expanding along the first row, \[ |A| = 2 \begin{vmatrix} \alpha & 7 \\ \beta & 3 \end{vmatrix} +3 \begin{vmatrix} 1 & 7 \\ 3 & 3 \end{vmatrix} +5 \begin{vmatrix} 1 & \alpha \\ 3 & \beta \end{vmatrix}. \] \[ = 2(3\alpha-7\beta) +3(3-21) +5(\beta-3\alpha). \] \[ = 6\alpha-14\beta-54+5\beta-15\alpha. \] \[ = -9\alpha-9\beta-54. \] \[ |A| = -9(\alpha+\beta+6). \]

Step 3: Use the condition for non-basis.
\[ -9(\alpha+\beta+6)=0. \] Hence, \[ \alpha+\beta+6=0. \] Replacing \((\alpha,\beta)\) by \((x,y)\), \[ x+y+6=0. \]