. For the matrix $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\1&1\end{array}} \right]$, find the numbers a and b such that ${A^2} + aA + bI = 0.$

We are given that, $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\1&1\end{array}} \right]$

Now, ${A^2} + aA + bI = O$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\1&1\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\1&1\end{array}} \right] + a\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2\\1&1\end{array}} \right] + b\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right] = O$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{9 + 2}&{6 + 2}\\{3 + 1}&{2 + 1}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3a}&{2a}\\a&a\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}b&0\\0&b\end{array}} \right] = O$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}&8\\4&3\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{3a + b}&{2a}\\a&{a + b}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&0\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11 + 3a + b}&{8 + 2a}\\{4 + a}&{3 + a + b}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&0\end{array}} \right]$

$\Rightarrow$ $4 + a = 0 \Rightarrow a = - 4$

Also, $3 + a + b = 0 \Rightarrow b = - 3 + 4 \Rightarrow b = 1$

Hence,$a = - 4,b = 1$