If A is `2xx2` matrix such that `A[{:(" "1),(-1):}]=[{:(-1),(2):}]and A^2[{:(" "1),(-1):}]=[{:(1),(0):}]` , then trace of A is (where the trace of the matrix is the sum of all principal diagonal elements of the matrix )

Options

A. 1
B. 0
C. 2
D. 5

Answer

To solve the problem, we need to find the trace of the matrix \( A \), which is defined as the sum of the principal diagonal elements of the matrix. We are given the following conditions:

1. \( A \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \)
2. \( A^2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)

Let's denote the matrix \( A \) as:

\[ A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \]

### Step 1: Use the first condition

From the first condition, we can write:

\[ A \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha - \beta \\ \gamma - \delta \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \]

From this, we can derive two equations:

1. \( \alpha - \beta = -1 \) (Equation 1)
2. \( \gamma - \delta = 2 \) (Equation 2)

### Step 2: Use the second condition

Next, we need to use the second condition. We can express \( A^2 \) as:

\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \]

Calculating \( A^2 \):

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

Now, we apply this to the vector \( \begin{pmatrix} 1 \\ -1 \end{pmatrix} \):

\[ A^2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 + \beta \gamma - (\alpha \beta + \beta \delta) \\ \gamma \alpha + \delta \gamma - (\gamma \beta + \delta^2) \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \]

This gives us two more equations:

1. \( \alpha^2 + \beta \gamma - \alpha \beta - \beta \delta = 1 \) (Equation 3)
2. \( \gamma \alpha + \delta \gamma - \gamma \beta - \delta^2 = 0 \) (Equation 4)

### Step 3: Solve the equations

From Equation 1, we can express \( \alpha \) in terms of \( \beta \):

\[ \alpha = \beta - 1 \]

From Equation 2, we can express \( \gamma \) in terms of \( \delta \):

\[ \gamma = \delta + 2 \]

Now, substituting \( \alpha \) and \( \gamma \) into Equations 3 and 4:

Substituting into Equation 3:

\[ (\beta - 1)^2 + \beta(\delta + 2) - (\beta - 1)\beta - \beta\delta = 1 \]

Expanding and simplifying:

\[ \beta^2 - 2\beta + 1 + \beta\delta + 2\beta - \beta^2 + \beta - \beta\delta = 1 \]

This simplifies to:

\[ 1 = 1 \]

This equation is always true, indicating that we have a relationship but no new information.

Now substituting into Equation 4:

\[ (\delta + 2)(\beta - 1) + \delta(\delta + 2) - \beta(\delta + 2) - \delta^2 = 0 \]

Expanding and simplifying:

\[ \delta\beta - \delta + 2\beta - 2 + \delta^2 + 2\delta - \beta\delta - \delta^2 = 0 \]

This simplifies to:

\[ \beta + \delta - 2 = 0 \implies \beta + \delta = 2 \]

### Step 4: Find values of \( \alpha \) and \( \delta \)

Now we have:

1. \( \alpha = \beta - 1 \)
2. \( \beta + \delta = 2 \)

Let \( \delta = 2 - \beta \).

Substituting \( \delta \) into Equation 2:

\[ \gamma = (2 - \beta) + 2 = 4 - \beta \]

### Step 5: Find the trace

Now we can find the trace of \( A \):

\[ \text{Trace}(A) = \alpha + \delta = (\beta - 1) + (2 - \beta) = 1 \]

### Final Answer

Thus, the trace of the matrix \( A \) is:

\[ \boxed{1} \]