Matrix A = $\begin{bmatrix$}[r] 2 & -2\ -2 & 0 \end{bmatrix} and matrix B = $\begin{bmatrix$}[r] 5 & 5\ 5 & 5 \end{bmatrix}

Statement 1 : AB = 0

Statement 2 : AB = 0, even if A ≠ 0 and B ≠ 0.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Given,

Matrix A = $\begin{bmatrix$}[r] 2 & -2\ -2 & 0 \end{bmatrix} and matrix B = $\begin{bmatrix$}[r] 5 & 5\ 5 & 5 \end{bmatrix}

$\Rightarrow AB = \begin{bmatrix$}[r] 2 & -2\ -2 & 0 \end{bmatrix}. \begin{bmatrix}[r] 5 & 5\ 5 & 5 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 2 \times 5 + (-2) \times 5 & 2 \times 5 + (-2) \times 5\ (-2) \times 5 + 0 \times 5 & (-2) \times 5 + 0 \times 5 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 10 - 10 & 10 - 10\ -10 + 0 & -10 + 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 0 & 0\ -10 & -10 \end{bmatrix}

So, AB ≠ 0

∴ Statement 1 is false.

Lets take $\text{ A } = \begin{bmatrix$}[r] 1 & -1\ 2 & -2 \end{bmatrix} \text{ and B } = \begin{bmatrix}[r] 3 & 3 \ 3 & 3 \end{bmatrix}

$AB = \begin{bmatrix$}[r] 1 & -1\ 2 & -2 \end{bmatrix} . \begin{bmatrix}[r] 3 & 3 \ 3 & 3 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 1 \times 3 + (-1) \times 3 & 1 \times 3 + (-1) \times 3\ 2 \times 3 + (-2) \times 3 & 2 \times 3 + (-2) \times 3\end{bmatrix} \\[1em] = \begin{bmatrix}[r] 3 - 3 & 3 - 3\ 6 - 6 & 6 - 6 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 0 & 0\ 0 & 0 \end{bmatrix}\\[1em]

Hence, it is proved that AB can be equal to 0, even if A ≠ 0 and B ≠ 0.

∴ Statement 2 is true.

Hence, option 4 is the correct option.