Question

If the square of the matrix $\begin{pmatrix} a & b \\ a & -a \end{pmatrix}$ is the unit matrix, then $b$ is equal to:

• A. \( \frac{a}{1+a^2} \)
• B. \( \frac{1-a^2}{a} \)
• C. \( \frac{1+a^2}{a} \)
• D. \( \frac{a}{1-a^2} \)
• E. \( 1+a^2 \)

Hint:

The square of a matrix \( \begin{pmatrix} a & b c & -a \end{pmatrix} \) is always a scalar matrix of the form \( (a^2+bc)I \). Setting this equal to \( I \) gives \( a^2+bc=1 \). In our case, \( c=a \), so \( a^2+ab=1 \).

Answer 1

The Correct Option is B

Concept: A unit matrix (or identity matrix) \( I \) for a \( 2 \times 2 \) system is \( \begin{pmatrix} 1 & 0 0 & 1 \end{pmatrix} \). We need to calculate \( M^2 \) and equate it to \( I \).

Step 1: Calculate the square of the given matrix.
Let \( M = \begin{pmatrix} a & b a & -a \end{pmatrix} \). \[ M^2 = \begin{pmatrix} a & b a & -a \end{pmatrix} \begin{pmatrix} a & b a & -a \end{pmatrix} \] Using matrix multiplication (row by column): \[ M^2 = \begin{pmatrix} (a \cdot a + b \cdot a) & (a \cdot b + b \cdot -a) (a \cdot a + -a \cdot a) & (a \cdot b + -a \cdot -a) \end{pmatrix} \] \[ M^2 = \begin{pmatrix} a^2 + ab & ab - ab a^2 - a^2 & ab + a^2 \end{pmatrix} = \begin{pmatrix} a^2 + ab & 0 0 & ab + a^2 \end{pmatrix} \]

Step 2: Equate \( M^2 \) to the unit matrix \( I \).
\[ \begin{pmatrix} a^2 + ab & 0 0 & ab + a^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 0 & 1 \end{pmatrix} \] From the equality of matrices, the corresponding elements must be equal: \[ a^2 + ab = 1 \]

Step 3: Solve for \( b \).
\[ ab = 1 - a^2 \] \[ b = \frac{1 - a^2}{a} \]