if for a matrix `A, A^2+I=O`, where I is the identity matrix, then A equals

Options

A. `{:[(1,0),(0,1)]:}`
B. `{:[(i,0),(0,-i)]:}`
C. `{:[(1,2),(-1,1)]:}`
D. `{:[(-1,0),(0,-1)]:}`

Answer

To solve the equation \( A^2 + I = O \) where \( I \) is the identity matrix and \( O \) is the zero matrix, we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ A^2 + I = O \]
Rearranging this gives:
\[ A^2 = -I \]

### Step 2: Understanding the Implications
The equation \( A^2 = -I \) implies that when we square the matrix \( A \), we get the negative of the identity matrix. This means that \( A \) must be a matrix whose eigenvalues are complex numbers, specifically \( i \) and \( -i \) (where \( i \) is the imaginary unit).

### Step 3: Considering Possible Forms of \( A \)
Since \( A^2 = -I \), we can consider a 2x2 matrix of the form:
\[ A = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \]
This matrix is a standard representation of a rotation in the complex plane.

### Step 4: Verifying the Solution
To verify that this matrix satisfies the original equation, we compute \( A^2 \):
\[ A^2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \]
Calculating this product:
\[ = \begin{pmatrix} 0 \cdot 0 + (-i) \cdot i & 0 \cdot (-i) + (-i) \cdot 0 \\ i \cdot 0 + 0 \cdot i & i \cdot (-i) + 0 \cdot 0 \end{pmatrix} \]
\[ = \begin{pmatrix} 0 + 1 & 0 \\ 0 & -(-1) \end{pmatrix} \]
\[ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]
Thus,
\[ A^2 = -I \implies A^2 + I = 0 \]

### Conclusion
The matrix \( A \) that satisfies the equation \( A^2 + I = O \) is:
\[ A = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \]