Prove A² = O

Prove \(A^2 = O\)

Question:
If \[ A=\begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \] show that: \[ A^2 = O \]

Solution:

Step 1: Compute \(A^2 = A \cdot A\)

\[ A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \]

Step 2: Multiply

\[ = \begin{bmatrix} ab\cdot ab + b^2(-a^2) & ab\cdot b^2 + b^2(-ab) \\ -a^2\cdot ab + (-ab)(-a^2) & -a^2\cdot b^2 + (-ab)(-ab) \end{bmatrix} \] \[ = \begin{bmatrix} a^2b^2 – a^2b^2 & ab^3 – ab^3 \\ -a^3b + a^3b & -a^2b^2 + a^2b^2 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Conclusion:

\[ A^2 = O \]

Hence proved.

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