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# If$A = \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right]$,prove that$(A - 2I)(A - 3I) = 0$

Last updated date: 08th Aug 2024
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Hint: $I = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
Given, $A = \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right]$. First, we’ll compute $(A - 2I)$where$I = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$.
$(A - 2I) \Rightarrow \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&2 \end{array}} \right] \\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {4 - 2}&{2 - 0} \\ { - 1 - 0}&{1 - 2} \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&2 \\ { - 1}&{ - 1} \end{array}} \right] \\$
Now, $(A - 3I) \Rightarrow \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} 4&2 \\ { - 1}&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&0 \\ 0&3 \end{array}} \right] \\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {4 - 3}&{2 - 0} \\ { - 1 - 0}&{1 - 3} \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2 \\ { - 1}&{ - 2} \end{array}} \right] \\$
And, $(A - 2I)(A - 3I) \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&2 \\ { - 1}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2 \\ { - 1}&{ - 2} \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} {2 \times 1 + 2 \times ( - 1)}&{2 \times 2 + 2 \times ( - 2)} \\ {( - 1) \times 1 + ( - 1) \times ( - 1)}&{( - 1) \times 2 + ( - 1) \times ( - 2)} \end{array}} \right] \\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2 - 2}&{4 - 4} \\ { - 1 + 1}&{ - 2 + 2} \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right] \Rightarrow 0 \\$