Question

Let $A$ be a $2 \times 2$ matrix such that $\text{trace}(A) = \det(A) = a (a \neq 0)$. Then the trace of $A^{-1}$ is}

• A. $a$
• B. 1
• C. $1/a$
• D. $-1/a$

Hint:

For any $2 \times 2$ matrix:
$\text{trace}(A^{-1}) = \frac{\text{trace}(A)}{\det(A)}$.
Since we are given $\text{trace}(A) = \det(A)$, their ratio is always 1, regardless of the value of $a$ (as long as $a \neq 0$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a $2 \times 2$ matrix $A$ where both the trace and the determinant are equal to a non-zero scalar $a$. We need to find the trace of its inverse matrix, $A^{-1}$.

Step 2: Key Formula or Approach:
Let the eigenvalues of matrix $A$ be $\lambda_1$ and $\lambda_2$.
- The trace of a matrix is the sum of its eigenvalues: $\text{trace}(A) = \lambda_1 + \lambda_2 = a$.
- The determinant of a matrix is the product of its eigenvalues: $\det(A) = \lambda_1 \lambda_2 = a$.
- The eigenvalues of the inverse matrix $A^{-1}$ are the reciprocals of the eigenvalues of $A$: $\frac{1}{\lambda_1}$ and $\frac{1}{\lambda_2}$.
- Therefore, the trace of $A^{-1}$ is the sum of its eigenvalues: $\text{trace}(A^{-1}) = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$.

Step 3: Detailed Explanation:

• Set up the expression for the trace of $A^{-1}$:
\[ \text{trace}(A^{-1}) = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]
• Find a common denominator to simplify the expression:
\[ \text{trace}(A^{-1}) = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2} \]
• Substitute the known values of the sum ($\text{trace}(A) = a$) and the product ($\det(A) = a$):
\[ \text{trace}(A^{-1}) = \frac{a}{a} \]
• Since $a \neq 0$, the expression simplifies to:
\[ \text{trace}(A^{-1}) = 1 \]

Step 4: Final Answer:
The trace of $A^{-1}$ is equal to 1.