Matrices & Determinants Question - None

Question

If $A = \begin{bmatrix}a & b\ 0 & a\end{bmatrix}$ is $n$th root of $I_2$, then choose the correct statements:
(i) if $n$ is odd, $a = 1,\ b = 0$
(ii) if $n$ is odd, $a = -1,\ b = 0$
(iii) if $n$ is even, $a = 1,\ b = 0$
(iv) if $n$ is even, $a = -1,\ b = 0$

Accepted Answer

Step 1: Set up the condition. Given A^n = I₂, we need:

$$\begin{bmatrix}a & b\ 0 & a\end{bmatrix}^n = \begin{bmatrix}1 & 0\ 0 & 1\end{bmatrix}$$

Step 2: Compute A² first.

$$A^2 = \begin{bmatrix}a & b\ 0 & a\end{bmatrix}\begin{bmatrix}a & b\ 0 & a\end{bmatrix} = \begin{bmatrix}a^2 & 2ab\ 0 & a^2\end{bmatrix}$$

Step 3: General formula for A^n. By induction, for upper triangular matrix with diagonal entries a:

$$A^n = \begin{bmatrix}a^n & nba^{n-1}\ 0 & a^n\end{bmatrix}$$

Step 4: Apply condition A^n = I₂. Comparing with identity:

$$a^n = 1 ext{ and } nba^{n-1} = 0$$

Step 5: Analyze a^n = 1. The solutions to a^n = 1 are the nth roots of unity. Real solutions:

• If n is odd: a^n = 1 has only a = 1 (since a^n is strictly monotonic for odd n in reals)

• If n is even: a^n = 1 gives a = 1 or a = -1

Step 6: Analyze nba^{n-1} = 0. Since n ≠ 0:

• If a ≠ 0, then a^{n-1} ≠ 0, so b = 0

• Since a^n = 1, we have a ≠ 0, thus b = 0 in all cases

Step 7: Verify each statement.

(i) n is odd, a = 1, b = 0: Check: A = I₂, so A^n = I₂ ✓ TRUE

(ii) n is odd, a = -1, b = 0: Check: A^n = (-1)^n I₂ = -I₂ ≠ I₂ (since n is odd) ✗ FALSE

(iii) n is even, a = 1, b = 0: Check: A^n = I₂ ✓ TRUE

(iv) n is even, a = -1, b = 0: Check: A^n = (-1)^n I₂ = I₂ (since n is even) ✓ TRUE

Step 8: Identify correct statements. Statements (i), (iii), and (iv) are correct.

∴ Answer: D

Question:

Step-by-Step Solution

Master Matrices & Determinants with Mathbee