Question

If \(A=\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\)and A = A^-1, then the value of x is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

To find the value of \(x\) such that \(A = A^{-1}\), where \(A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\), follow these steps:

1. For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is given by \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \).

2. For \( A \) to be equal to its inverse \( A^{-1} \), we first need to calculate \( A^{-1} \):

\( \det(A) = x \cdot 0 - 1 \cdot 1 = -1 \)

\( A^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 \\ -1 & x \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix} \)

3. Since \( A = A^{-1} \), compare elements:

\( \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix} \)

Comparing first row, first column elements: \( x = 0 \)

Comparing second row, second column elements: \( 0 = -x \) confirms \( x = 0 \)

4. Thus, the value of \( x \) is 0.