If `[alphabetagamma-alpha]` is to be square root of two-rowed unit matrix, then `alpha,betaa n dgamma` should satisfy the relation. `1-alpha^2+betagamma=0` b. `alpha^2+betagamma=0` c. `1+alpha^2+betagamma=0` d. `1-alpha^2-betagamma=0`

Options

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

Explanation

To solve the problem, we need to analyze the given matrix and its relationship with the square root of the two-rowed unit matrix. Let's denote the matrix as follows:

\[ M = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

We know that \( M \) is equal to the square root of the two-rowed unit matrix, which can be represented as:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Taking the square of both sides, we have:

\[ M^2 = I \]

Now, we will compute \( M^2 \):

\[ M^2 = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

Calculating the product:

1. The element at (1,1):
\[ \alpha \cdot \alpha + \beta \cdot \gamma = \alpha^2 + \beta\gamma \]

2. The element at (1,2):
\[ \alpha \cdot \beta + \beta \cdot (-\alpha) = \alpha\beta - \alpha\beta = 0 \]

3. The element at (2,1):
\[ \gamma \cdot \alpha + (-\alpha) \cdot \gamma = \gamma\alpha - \alpha\gamma = 0 \]

4. The element at (2,2):
\[ \gamma \cdot \beta + (-\alpha) \cdot (-\alpha) = \gamma\beta + \alpha^2 \]

Thus, we can write:

\[ M^2 = \begin{pmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \gamma\beta + \alpha^2 \end{pmatrix} \]

Setting \( M^2 \) equal to the identity matrix \( I \):

\[ \begin{pmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \gamma\beta + \alpha^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we can equate the corresponding elements:

1. From the (1,1) position:
\[ \alpha^2 + \beta\gamma = 1 \]

2. From the (2,2) position:
\[ \gamma\beta + \alpha^2 = 1 \]

Both equations yield the same relation. Thus, we can conclude that:

\[ \alpha^2 + \beta\gamma = 1 \]

Rearranging gives:

\[ 1 - \alpha^2 - \beta\gamma = 0 \]

Now, we can identify the correct relation from the options given:

The correct relation is:
**d. \( 1 - \alpha^2 - \beta\gamma = 0 \)**