Question

Let \[ A= \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}. \] Then the determinant of \(A^2\) is:

• A. \(9\)
• B. \(16\)
• C. \(25\)
• D. \(36\)

Hint:

Remember the determinant rule \[ |A^n|=(|A|)^n. \] Whenever powers of a matrix appear inside a determinant, use this property first before attempting matrix multiplication.

Answer 1

The Correct Option is A

Concept: One of the most important properties of determinants is \[ |AB|=|A||B|. \] As a consequence, \[ |A^2| = |A|^2. \] Therefore, instead of calculating the matrix \(A^2\) first, we can directly compute the determinant of \(A\) and then square it. This significantly reduces the amount of computation required.

Step 1: Write the given matrix. \[ A= \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}. \]

Step 2: Find the determinant of \(A\). For a \(2\times2\) matrix \[ \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \] the determinant is \[ ad-bc. \] Therefore, \[ |A| = (1)(1)-(2)(2). \] \[ = 1-4. \] \[ =-3. \]

Step 3: Use the determinant property. Since \[ |A^2| = |A|^2, \] we obtain \[ |A^2| = (-3)^2. \] \[ =9. \]

Step 4: Verification by direct multiplication. Computing \[ A^2 = \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}, \] gives \[ A^2= \begin{bmatrix} 5 & 4\\ 4 & 5 \end{bmatrix}. \] Now, \[ |A^2| = (5)(5)-(4)(4). \] \[ = 25-16. \] \[ =9. \] Thus our answer is verified.

Step 5: Final Conclusion. \[ \boxed{|A^2|=9} \] Hence the correct answer is \[ \boxed{\text{Option (A)}}. \]