Question

If \(\begin{bmatrix} \alpha & \beta \gamma & -\alpha \end{bmatrix}\) is to be the square root of the two-rowed unit matrix, then \(\alpha, \beta\) and \(\gamma\) should satisfy the relation

• A. \(1 + \alpha^2 + \beta\gamma = 0\)
• B. \(1 - \alpha^2 - \beta\gamma = 0\)
• C. \(1 - \alpha^2 + \beta\gamma = 0\)
• D. \(\alpha^2 + \beta\gamma - 1 = 0\)

Hint:

Matrix square roots require careful multiplication, especially when off-diagonals may not vanish automatically.

Answer 1

The Correct Option is B

To determine the relation that \(\alpha\), \(\beta\), and \(\gamma\) must satisfy for the matrix \(\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\) to be the square root of the two-rowed unit matrix \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), we need to carry out the following steps:

1. Matrix Multiplication: The matrix \(\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\) should satisfy \(\left(\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\right)^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\). Let's calculate the square of the given matrix:

\(\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta \gamma & \alpha \beta - \alpha \beta \\ \gamma \alpha - \alpha \gamma & \gamma \beta + \alpha^2 \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \beta \gamma \end{bmatrix}\)

1. Setting Equal to Identity Matrix: For this result to equal the identity matrix \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), we need: \(\alpha^2 + \beta \gamma = 1\).
2. Conclusion: The resultant equation to be satisfied is: \(1 - \alpha^2 - \beta \gamma = 0\).

Thus, the correct answer is: \(1 - \alpha^2 - \beta\gamma = 0\).