If `{:[(alpha,beta),(gamma,-alpha)]:}` is to be the square root of two-rowed unit matrix, then `alpha,beta and gamma` should satisfy the relation

Options

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`

Answer

To solve the problem, we need to find the relationship that the variables \( \alpha \), \( \beta \), and \( \gamma \) must satisfy if the matrix

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

is the square root of the two-rowed unit matrix, which is

\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \]

### Step 1: Square the given matrix

We start by squaring the matrix:

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

### Step 2: Perform matrix multiplication

Using the rules of matrix multiplication, we compute:

\[ \begin{pmatrix} \alpha \cdot \alpha + \beta \cdot \gamma & \alpha \cdot \beta + \beta \cdot (-\alpha) \\ \gamma \cdot \alpha + (-\alpha) \cdot \gamma & \gamma \cdot \beta + (-\alpha) \cdot (-\alpha) \end{pmatrix} \]

This simplifies to:

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

### Step 3: Set the result equal to the unit matrix

Since we want this result to equal the unit matrix, we have:

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

### Step 4: Set up the equations

From the matrix equality, we can derive the following equation:

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

### Step 5: Rearranging the equation

Rearranging gives us:

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

### Conclusion

Thus, the relationship that \( \alpha \), \( \beta \), and \( \gamma \) must satisfy is:

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