Question

If \(A=\begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}\), then the determinant of \((A^{2}-2A)^{-1}\) is...

• A. \(\frac{1}{5}\)
• B. \(25\)
• C. \(\frac{1}{25}\)
• D. \(-25\)

Hint:

Factor the expression as A times (A minus 2I) and use the product and inverse rules for determinants.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For square matrices, \(|XY|=|X||Y|\) and \(|X^{-1}|=\frac{1}{|X|}\).

Step 2: Find |A|
\(|A|=1\cdot1-3\cdot2=-5\).

Step 3: Factor the expression
\(A^{2}-2A=A(A-2I)\) with \(A-2I=\begin{bmatrix}-1&3\\2&-1\end{bmatrix}\).
\(|A-2I|=1-6=-5\).

Step 4: Find the required determinant
\(|A^{2}-2A|=(-5)(-5)=25\).
Therefore \(|(A^{2}-2A)^{-1}|=\frac{1}{25}\).

Final Answer:
The determinant of the inverse is one twenty-fifth. \[ \boxed{\text{(C) }\dfrac{1}{25}} \]