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Mathematics

Matrix A = [2−2−22]\begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} and matrix B = [5555]\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.

Matrices

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Answer

Given,

Matrix A = [2−2−22]\begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} and matrix B = [5555]\begin{bmatrix}[r] 5 & 5 \ 5 & 5 \end{bmatrix}

Solving,

⇒AB=[2−2−22].[5555]=[2×5+(−2)×52×5+(−2)×5(−2)×5+2×5(−2)×5+2×5]=[10−1010−10−10+10−10+10]=[0000]\Rightarrow AB = \begin{bmatrix}[r] 2 & -2 \ -2 & 2 \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 + 2 \times 5 & (-2) \times 5 + 2 \times 5 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 10 - 10 & 10 - 10 \ -10 + 10 & -10 + 10 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}

So, AB = 0 and also from above example it is proved that AB can be equal to zero even if A ≠ 0 and B ≠ 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

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