Question:
If \(\alpha\) and \(\beta\) are the roots of the equation \(1 + x + x^2 = 0\), then the matrix product \(\begin{bmatrix} 1 & \beta \\ \alpha & \alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ 1 & \beta \end{bmatrix}\) is equal to:
यदि \(\alpha\) और \(\beta\) समीकरण \(1 + x + x^2 = 0\) के मूल हैं, तो आव्यूह गुणनफल \(\begin{bmatrix} 1 & \beta \\ \alpha & \alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ 1 & \beta \end{bmatrix}\) किसके बराबर है?
Options:
- \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & -1 \\ -1 & -2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\)
- \(\begin{bmatrix} -1 & -1 \\ -1 & -2 \end{bmatrix}\)
Correct Answer: b)
Explanation:
The roots of \(x^2 + x + 1 = 0\) are the complex cube roots of unity: \(\alpha = \omega\) and \(\beta = \omega^2\).
We have \(\alpha + \beta = \omega + \omega^2 = -1\) and \(\alpha\beta = \omega^3 = 1\).
Multiplying the matrices:
\[\begin{bmatrix} 1 & \beta \\ \alpha & \alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ 1 & \beta \end{bmatrix} = \begin{bmatrix} \alpha + \beta & \beta + \beta^2 \\ \alpha^2 + \alpha & \alpha\beta + \alpha\beta \end{bmatrix}\]
– \((1, 1)\): \(\alpha + \beta = -1\)
– \((1, 2)\): \(\beta + \beta^2 = \omega^2 + \omega^4 = \omega^2 + \omega = -1\)
– \((2, 1)\): \(\alpha^2 + \alpha = \omega^2 + \omega = -1\)
– \((2, 2)\): \(2\alpha\beta = 2(1) = 2\)
Thus, the product matrix is:
\[\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix}\]
Hence, the correct option is (b).
\(x^2 + x + 1 = 0\) के मूल इकाई के घनमूल \(\alpha = \omega\) तथा \(\beta = \omega^2\) हैं।
\(\alpha + \beta = -1\) तथा \(\alpha\beta = 1\)।
आव्यूह गुणन:
\[\begin{bmatrix} 1 & \beta \\ \alpha & \alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ 1 & \beta \end{bmatrix} = \begin{bmatrix} \alpha + \beta & \beta(1 + \beta) \\ \alpha(\alpha + 1) & 2\alpha\beta \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix}\]
अतः सही विकल्प (b) है।