Question

The rank of the matrix $A$ given below is one. The ratio $\dfrac{\alpha}{\beta}$ is ______ (rounded off to the nearest integer). \[ A = \begin{bmatrix} 1 & 4 \\ -3 & \alpha \\ \beta & 6 \end{bmatrix} \]

Hint:

For rank $=1$, all $2 \times 2$ minors of the matrix must be zero. Use this property to solve systematically.

Answer 1

Correct Answer: -8

Rank $=1 \Rightarrow$ all $2 \times 2$ minors must vanish.
Minor from rows 1 and 2: \[ \det \begin{bmatrix} 1 & 4 \\ -3 & \alpha \end{bmatrix} = 1 . \alpha - (-3) . 4 = \alpha + 12 = 0 \;\Rightarrow\; \alpha = -12. \] Minor from rows 1 and 3: \[ \det \begin{bmatrix} 1 & 4 \\ \beta & 6 \end{bmatrix} = 1 . 6 - \beta . 4 = 6 - 4\beta = 0 \;\Rightarrow\; \beta = 1.5. \] Thus \[ \frac{\alpha}{\beta} = \frac{-12}{1.5} = -8. \]