Prove \((A – 2I)(A – 3I) = O\)

Solution:

Step 1: Identity Matrix

\[ I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \]

Step 2: Compute \(A – 2I\)

\[ A – 2I = \begin{bmatrix} 4-2 & 2 \\ -1 & 1-2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \]

Step 3: Compute \(A – 3I\)

\[ A – 3I = \begin{bmatrix} 4-3 & 2 \\ -1 & 1-3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \]

Step 4: Multiply

\[ (A – 2I)(A – 3I) = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \] \[ = \begin{bmatrix} 2(1)+2(-1) & 2(2)+2(-2) \\ -1(1)+(-1)(-1) & -1(2)+(-1)(-2) \end{bmatrix} \] \[ = \begin{bmatrix} 2-2 & 4-4 \\ -1+1 & -2+2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Conclusion:

\[ (A – 2I)(A – 3I) = O \]

Hence proved.