Question

Let $A = \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix}$ and I be unit matrix of order 2.

Statement-1: $(I+A)^{532} = I + 532A$ and

Statement-2: If B be a nilpotent matrix of index 2 and I be a unit matrix of same order as B then $(I+B)^n = I + nB, n \in N$.

• A. Statement-1 is True, Statement-2 is True
• B. Statement-1 is False, Statement-2 is False
• C. Statement-1 is True, Statement-2 is False.
• D. Statement-1 is False, Statement-2 is True.

Answer 1

Answer: (A)

Statement-1 is True, Statement-2 is True

Answer 2

Statement-1:

We are given the matrix $A = \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix}$ and the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. We need to check if $(I+A)^{532} = I + 532A$.

First, let's calculate $I+A$: $I+A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}$.

Let's calculate the powers of $(I+A)$: $(I+A)^1 = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}$ $(I+A)^2 = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix}$ $(I+A)^3 = \begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -9 \\ 0 & 1 \end{bmatrix}$

By observing the pattern, it appears that $(I+A)^n = \begin{bmatrix} 1 & -3n \\ 0 & 1 \end{bmatrix}$ for any positive integer $n$.

For $n=532$, we have $(I+A)^{532} = \begin{bmatrix} 1 & -3 \cdot 532 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1596 \\ 0 & 1 \end{bmatrix}$.

Now let's calculate $I + 532A$: $I + 532A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 532 \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -1596 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1596 \\ 0 & 1 \end{bmatrix}$.

Since $(I+A)^{532} = \begin{bmatrix} 1 & -1596 \\ 0 & 1 \end{bmatrix}$ and $I + 532A = \begin{bmatrix} 1 & -1596 \\ 0 & 1 \end{bmatrix}$, Statement-1 is True.

Statement-2:

We are given that B is a nilpotent matrix of index 2, which means $B \neq 0$ and $B^2 = 0$. We need to check if $(I+B)^n = I + nB$ for $n \in N$.

Since I is the identity matrix, it commutes with any matrix B of the same order, i.e., $IB = BI$. Therefore, we can use the binomial expansion for matrices:

$(I+B)^n = \binom{n}{0} I^n B^0 + \binom{n}{1} I^{n-1} B^1 + \binom{n}{2} I^{n-2} B^2 + \binom{n}{3} I^{n-3} B^3 + \dots + \binom{n}{n} I^0 B^n$.

Since $B^2 = 0$, it follows that $B^3 = B^2 B = 0 \cdot B = 0$, $B^4 = B^3 B = 0 \cdot B = 0$, and so on. Thus, $B^k = 0$ for all $k \ge 2$.

The binomial expansion simplifies to: $(I+B)^n = \binom{n}{0} I + \binom{n}{1} B + \binom{n}{2} \cdot 0 + \binom{n}{3} \cdot 0 + \dots + \binom{n}{n} \cdot 0$ $(I+B)^n = 1 \cdot I + n \cdot B$ $(I+B)^n = I + nB$.

This formula holds for any natural number $n$. Therefore, Statement-2 is True.

Both Statement-1 and Statement-2 are true. Note that the matrix A in Statement-1 is a nilpotent matrix of index 2, since $A^2 = \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & -3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ and $A \neq 0$. Statement-1 is a specific instance of the general result given in Statement-2, for $B=A$ and $n=532$.