Question

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

Answer 1

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, \[
  (A - 2I)(A - 3I) \Rightarrow \left[ {\begin{array}{*{20}{c}}
  2&2 \\
  { - 1}&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&{ - 2}
\end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {2 \times 1 + 2 \times ( - 1)}&{2 \times 2 + 2 \times ( - 2)} \\
  {( - 1) \times 1 + ( - 1) \times ( - 1)}&{( - 1) \times 2 + ( - 1) \times ( - 2)}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {2 - 2}&{4 - 4} \\
  { - 1 + 1}&{ - 2 + 2}
\end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right] \Rightarrow 0 \\
\]
Hence Proved.

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