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 find the trace of the matrix \( A \), we will start by defining the matrix \( A \) as follows:

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

Given the conditions in the problem, we have:
1. \( A \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \)
2. \( A^2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)

### Step 1: Solve the first condition

From the first condition:
\[ A \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha - \beta \\ \gamma - \delta \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \]

This gives us two equations:
1. \( \alpha - \beta = -1 \) (Equation 1)
2. \( \gamma - \delta = 2 \) (Equation 2)

### Step 2: Solve the second condition

Now, we need to calculate \( A^2 \) and apply the second condition:
\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} = \begin{pmatrix} \alpha^2 + \beta \gamma & \alpha \beta + \beta \delta \\ \gamma \alpha + \delta \gamma & \gamma \beta + \delta^2 \end{pmatrix} \]

Now applying the second condition:
\[ A^2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 + \beta \gamma & \alpha \beta + \beta \delta \\ \gamma \alpha + \delta \gamma & \gamma \beta + \delta^2 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 + \beta \gamma - (\alpha \beta + \beta \delta) \\ \gamma \alpha + \delta \gamma - (\gamma \beta + \delta^2) \end{pmatrix} \]

This should equal:
\[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \]

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: Substitute and solve

From Equation 1, we have \( \alpha = \beta - 1 \). Substitute \( \alpha \) into Equation 3:
\[ (\beta - 1)^2 + \beta \gamma - (\beta - 1) \beta - \beta \delta = 1 \]
Expanding this:
\[ \beta^2 - 2\beta + 1 + \beta \gamma - \beta^2 + \beta - \beta \delta = 1 \]
Simplifying:
\[ \beta \gamma - \beta \delta - \beta + 1 = 1 \implies \beta \gamma - \beta \delta - \beta = 0 \implies \beta (\gamma - \delta - 1) = 0 \]

This gives us two cases:
1. \( \beta = 0 \)
2. \( \gamma - \delta - 1 = 0 \implies \gamma = \delta + 1 \)

### Step 4: Solve for \( \beta = 0 \)

If \( \beta = 0 \):
From Equation 1: \( \alpha = -1 \)

From Equation 2: \( \gamma - \delta = 2 \implies \gamma = \delta + 2 \)

Substituting into Equation 4:
\[ 0 + \delta(\delta + 2) - 0 - \delta^2 = 0 \implies \delta^2 + 2\delta - \delta^2 = 0 \implies 2\delta = 0 \implies \delta = 0 \]
Thus, \( \gamma = 2 \).

### Step 5: Find the trace of \( A \)

Now we have:
\[ A = \begin{pmatrix} -1 & 0 \\ 2 & 0 \end{pmatrix} \]
The trace of \( A \) is:
\[ \text{Trace}(A) = \alpha + \delta = -1 + 0 = -1 \]

### Final Answer

The trace of \( A \) is \( -1 \).