Question

If $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\mathrm{and}I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],$ then
(i) find λ, μ so that A² = λA + μI
(ii) prove that A³ − 4A² + A = O

Answer 1

$$\begin{gathered} \left(i\right) \\ \mathrm{Given}:A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4+3& 6+6\\ 2+2& 3+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right] \\ {A}^2=\lambda A+\mu I \\ \Rightarrow \left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]=\lambda \left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]+\mu \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]=\left[\begin{array}{cc}2\lambda & 3\lambda \\ \lambda & 2\lambda \end{array}\right]+\left[\begin{array}{cc}\mu & 0\\ 0& \mu \end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]=\left[\begin{array}{cc}2\lambda +\mu & 3\lambda +0\\ \lambda +0& 2\lambda +\mu \end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}7& 12\\ 4& 7\end{array}\right]=\left[\begin{array}{cc}2\lambda +\mu & 3\lambda \\ \lambda & 2\lambda +\mu \end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{equal}. \\ \therefore 7=2\lambda +\mu ...\left(1\right) \\ 12=3\lambda \\ \Rightarrow \lambda =\frac{12}{3}=4 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\lambda \mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ 7=2\left(4\right)+\mu \\ \Rightarrow 7-8=\mu \\ \therefore \mu =-1 \end{gathered}$$

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