Question

If$A=\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$,prove that$(A-2I)(A-3I)=0$

Answer 1

Hint: $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Given, $A=\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$. First, we’ll compute $(A-2I)$where$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.
$$\begin{gathered} (A-2I)\Rightarrow \left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]-2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}4-2& 2-0\\ -1-0& 1-2\end{array}\right]\Rightarrow \left[\begin{array}{cc}2& 2\\ -1& -1\end{array}\right] \end{gathered}$$
Now, $$\begin{gathered} (A-3I)\Rightarrow \left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]-3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}4-3& 2-0\\ -1-0& 1-3\end{array}\right]\Rightarrow \left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right] \end{gathered}$$
And, $$\begin{gathered} (A-2I)(A-3I)\Rightarrow \left[\begin{array}{cc}2& 2\\ -1& -1\end{array}\right]\left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right]\Rightarrow \left[\begin{array}{cc}2\times 1+2\times (-1)& 2\times 2+2\times (-2)\\ (-1)\times 1+(-1)\times (-1)& (-1)\times 2+(-1)\times (-2)\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2-2& 4-4\\ -1+1& -2+2\end{array}\right]\Rightarrow \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\Rightarrow 0 \end{gathered}$$
Hence Proved.

Note: It is crucial to perform scalar multiplication with matrix and matrix addition/subtraction with accuracy to achieve the correct solution.