Question

For \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), if \( A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \) be such that \( A^2 = I \), then :

• A. \( 1 + a^2 + bc = 0 \)
• B. \( 1 - a^2 - bc = 0 \)
• C. \( 1 - a^2 + bc = 0 \)
• D. \( 1 + a^2 - bc = 0 \)

Hint:

For any matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the Cayley-Hamilton theorem states \( A^2 - (Tr(A))A + |A|I = 0 \).
If \( Tr(A) = 0 \), then \( A^2 = -|A|I \).
For \( A^2 = I \), we need \( -|A| = 1 \), which means \( |A| = -1 \).
The determinant \( |A| = -a^2 - bc \). Setting \( -a^2 - bc = -1 \) gives \( a^2 + bc = 1 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem explores the properties of square matrices, specifically focusing on a special type known as an involutory matrix. An involutory matrix \( A \) is defined by the property that its square is equal to the identity matrix \( I \). In the context of linear transformations, this means that the transformation represented by \( A \) is its own inverse; applying the transformation twice returns the system to its original state.
The identity matrix \( I \) of order \( 2 \times 2 \) is given as \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). We are provided with a general form for matrix \( A \) where the diagonal elements are \( a \) and \( -a \). This structure is significant because the sum of the diagonal elements (the trace of the matrix) is zero. Matrices with a trace of zero often play a central role in linear algebra and physics. The objective is to identify the algebraic relationship between the variables \( a, b, \) and \( c \) that makes the matrix equation \( A^2 = I \) true.

Step 2: Key Formula or Approach:
To solve this, we utilize the definition of matrix multiplication. For two matrices to be multiplied, the number of columns in the first must match the number of rows in the second. Since \( A \) is a \( 2 \times 2 \) matrix, \( A^2 = A \times A \) is also a \( 2 \times 2 \) matrix.
The general formula for the square of a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} x & y \\ z & w \end{bmatrix} \) is:
\[ A^2 = \begin{bmatrix} x^2 + yz & xy + yw \\ zx + wz & zy + w^2 \end{bmatrix} \]
Once the product is computed, we use the principle of matrix equality, which states that two matrices are equal if and only if every corresponding entry is identical. This will yield a system of equations.

Step 3: Detailed Explanation:
Let's calculate the square of matrix \( A \).
Given \( A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \).
The operation is \( A \cdot A \):
\[ A^2 = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \]
Calculating the first row, first column entry (\( R_1C_1 \)):
\[ (a)(a) + (b)(c) = a^2 + bc \]
Calculating the first row, second column entry (\( R_1C_2 \)):
\[ (a)(b) + (b)(-a) = ab - ab = 0 \]
Calculating the second row, first column entry (\( R_2C_1 \)):
\[ (c)(a) + (-a)(c) = ac - ac = 0 \]
Calculating the second row, second column entry (\( R_2C_2 \)):
\[ (c)(b) + (-a)(-a) = bc + a^2 \]
Assembling these results into a matrix:
\[ A^2 = \begin{bmatrix} a^2 + bc & 0 \\ 0 & bc + a^2 \end{bmatrix} \]
We are given the condition \( A^2 = I \). Therefore:
\[ \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Equating the top-left (and bottom-right) entries gives:
\[ a^2 + bc = 1 \]
This is a standard relationship. To match the options provided in the question, we need to manipulate the equation. By subtracting both \( a^2 \) and \( bc \) from both sides, or moving everything to one side relative to the number 1, we get:
\[ 1 - a^2 - bc = 0 \]
This matches Option (2) exactly. We should also note that the off-diagonal entries are already 0, which satisfies the identity matrix structure regardless of the values of \( a, b, \) or \( c \).

Step 4: Final Answer:
The algebraic condition derived from the matrix squaring process is \( a^2 + bc = 1 \). Rearranging this leads to the expression \( 1 - a^2 - bc = 0 \). Consequently, the correct choice is option (2).