Chapter 8, Problem 5PS

(a)

To determine

To prove: $A^2-2A+5I=O$

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Proof:

Left Hand Side

  $\begin{array}{l}=A^2-2A+5I\\ =\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]\times \left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]-2\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]+5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ =\left[\begin{array}{cc}(1\cdot 1)+(2\cdot -2)& (1\cdot 2)+(2\cdot 1)\\ (-2\cdot 1)+(1\cdot -2)& (-2\cdot 2)+(1\cdot 1)\end{array}\right]-\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]+\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]\end{array}$

  $\begin{array}{l}=\left[\begin{array}{cc}-3& 4\\ -4& -3\end{array}\right]-\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]+\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]\\ =\left[\begin{array}{cc}(-3-2+5)& (4-4+0)\\ (-4+4+0)& (-3-2+5)\end{array}\right]\\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ =O\end{array}$

=Right Hand Side

Hence proved, $A^2-2A+5I=O$

(b)

To determine

To prove: $A^{-1}=\frac{1}{5}(2I-A)$

Explanation of Solution

Given: $A=\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Proof:

  $\left|A\right|=\left|\begin{array}{cc}1& 2\\ -2& 1\end{array}\right|=(1\cdot 1)-(2\cdot -2)=1+4=5$

The co-factors of the elements 1 and 2 of the first row in $\left|A\right|$ are 1 and 2 respectively, and the co-factors of the elements (-2) and 1 of the second row of $\left|A\right|$ are (-2) and 1 respectively.

Then, $Adj\cdot A={\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]}^T=\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]$

Now, Left Hand Side

  $\begin{array}{l}=A^{-1}\\ =\frac{Adj\cdot A}{\left|A\right|}\\ =\frac{1}{5}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\mathrm{......................}(1)\end{array}$

Right Hand Side

  $\begin{array}{l}=\frac{1}{5}(2I-A)\\ =\frac{1}{5}(2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right])\\ =\frac{1}{5}(\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]-\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right])\\ =\frac{1}{5}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\mathrm{...............................}(2)\end{array}$

From (1) and (2) we get,

Left Hand Side=Right Hand Side

Hence proved, $A^{-1}=\frac{1}{5}(2I-A)$

(c)

To determine

To prove:For any square matrix satisfying $A^2-2A+5I=O$ the inverse of A is given by $A^{-1}=\frac{1}{5}(2I-A)$.

Explanation of Solution

Given:Any square matrix A satisfies $A^2-2A+5I=O$ .

Proof:

  $\begin{array}{l}{A}^2-2A+5I=O\\ \Rightarrow A^2-2AI+5I=O[\because AI=A]\\ \Rightarrow 2AI-A^2=5I\\ \Rightarrow A\cdot (2I-A)=5I\\ \Rightarrow A\cdot \frac{1}{5}(2I-A)=I\end{array}$

Hence, by definition of inverse of A, that is, $A\cdot A^{-1}=I$ ,

  $A^{-1}=\frac{1}{5}(2I-A)$

Thus, for any square matrix satisfying $A^2-2A+5I=O$ the inverse of A is given by $A^{-1}=\frac{1}{5}(2I-A)$.