Determinants — Maths STD 12 Science — Question

$\therefore {A^2} = A.A = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$

$= \left[ {\begin{array}{*{20}{c}} {9 + 2}&{6 + 2} \\ {3 + 1}&{2 + 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right]$

$\therefore {A^2} + aA + b{I_2} = 0$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right] + a\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right] + b\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = 0$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {3a}&{2a} \\ a&a \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} b&0 \\ 0&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11 + 3a + b}&{8 + 2a + 0} \\ {4 + a + 0}&{3 + a + b} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$

$\therefore$ We have $11 + 3a + b = 0$….(i)

$8 + 2a + 0 = 0$ ...(ii)

$\Rightarrow 2a = - 8$

$\Rightarrow a = - 4$

Here a = -4 satisfies 4 + a + 0 = 0 also, therefore a = -4

Putting a = -4 in eq. (i), 11 - 12 + b = 0 $ \Rightarrow b - 1 = 0 \Rightarrow b = 1$

Here also b = 1 satisfies 3 + a + b = 0, therefore b = 1

Therefore, a = - 4 and b = 1