Question

(i) If $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$, show that A³ − 4A² + A = O.

(ii) If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that A² − 5A + 7I = O use this to find A⁴.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Given}:A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4+3& 6+6\\ 2+2& 3+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right] \\ {A}^3=A^{2}A \\ \Rightarrow A^3=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}14+12& 21+24\\ 8+7& 12+14\end{array}\right] \\ \Rightarrow A^3=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right] \\ {A}^3-4A^2+A \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right]-4\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26& 45\\ 15& 26\end{array}\right]-\left[\begin{array}{cc}28& 48\\ 16& 28\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}26-28+2& 45-48+3\\ 15-16+1& 26-28+2\end{array}\right] \\ \Rightarrow A^3-4A^2+A=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(ii\right) \\ \mathrm{Given}:A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right] \\ {A}^2-5A+7I \\ \Rightarrow A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right] \\ \Rightarrow A^2-5A+7I=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right] \\ \Rightarrow A^2-5A+7I=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0 \\ \mathrm{Hence}\mathrm{proved}. \\ \mathrm{Given}:A^2-5A+7I=0 \\ \Rightarrow A^2=5A-7I...\left(1\right) \\ \Rightarrow A^3=A\left(5A-7I\right)\left(\mathrm{Multilpying}\mathrm{by}A\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow A^3=5A^2-7AI \\ \Rightarrow A^3=5\left(5A-7I\right)-7A\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow A^3=25A-35I-7A \\ \Rightarrow A^3=18A-35I \\ \Rightarrow A^4=\left(18A-35I\right)A\left(\mathrm{Multilpying}\mathrm{by}A\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow A^4=18A^2-35A \\ \Rightarrow A^4=18\left(5A-7I\right)-35A\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow A^4=90A-126I-35A \\ \Rightarrow A^4=55A-126I \\ \Rightarrow A^4=55\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]-126\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow A^4=\left[\begin{array}{cc}165& 55\\ -55& 110\end{array}\right]-\left[\begin{array}{cc}126& 0\\ 0& 126\end{array}\right] \\ \Rightarrow A^4=\left[\begin{array}{cc}165-126& 55-0\\ -55-0& 110-126\end{array}\right] \\ \Rightarrow A^4=\left[\begin{array}{cc}39& 55\\ -55& -16\end{array}\right] \end{gathered}$$