Question

If \(\Delta = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (1 + ax^3)^b\), then which of the following statements are TRUE?
A. \(a = -1\) and \(b = 2\)
B. \(x = 1\) is a multiple root of \(\Delta = 0\)
C. \(x = 1\) is a simple root of \(\Delta = 0\)
D. \(x = 3\) is a simple root of \(\Delta = 0\)
Choose the correct answer from the options given below:

• A. A and D only
• B. A and C only
• C. A, B and D only
• D. A and B only

Hint:

Expand the determinant to get \((1 - x^3)^2\). Then compare with \((1 + ax^3)^b\) and factor \(1 - x^3\).

Answer 1

The Correct Option is D

Step 1: Spot the pattern:
The rows are cyclic shifts of \((1, x, x^2)\). We expand along the first column this time, instead of the first row, and then read off \(a\) and \(b\).

Step 2: Expand along column 1:
The first column is \((1, x^2, x)\).
The three minors are \(\begin{vmatrix} 1 & x \\ x^2 & 1 \end{vmatrix} = 1 - x^3\), \(\begin{vmatrix} x & x^2 \\ x^2 & 1 \end{vmatrix} = x - x^4\) and \(\begin{vmatrix} x & x^2 \\ 1 & x \end{vmatrix} = 0\). The signs alternate as +, -, +.
\[ \Delta = (1 - x^3) - x^2(x - x^4) + 0 = 1 - x^3 - x^3 + x^6 = (1 - x^3)^2 \]

Step 3: Read off a and b:
Since \(\Delta = (1 + (-1)x^3)^2\), we get \(a = -1\) and \(b = 2\). Statement A is TRUE.

Step 4: Look at the root \(x = 1\):
\(1 - x^3 = (1-x)(1+x+x^2)\). The determinant is the square of this, so the root \(x = 1\) is repeated twice. That is a multiple root, so B is TRUE and C is FALSE.

Step 5: Look at \(x = 3\):
At \(x = 3\), \(1 - x^3 = -26\), so \(\Delta = 676\), not zero. So D is FALSE.

Final Answer:
A and B are correct, which is option 4. \[ \boxed{\text{A and B only}} \]