Chapter 2.5, Problem 2P

Explanation of Solution

Finding out the Inverse:

Consider the following matrix,

$A=\left[\begin{array}{l}1\text{ 0 1}\\ 4\text{ 1 -2}\\ 3\text{ 1 -1}\end{array}\right]$

Suppose the inverse of this matrix is as follows:

$A^{-1}=\left[\begin{array}{l}\text{e f g}\\ \text{h i j}\\ \text{k }l\text{ m}\end{array}\right]$

Then we have to solve the following equations,

$\left[\begin{array}{l}1\text{ 0 1}\\ 4\text{ 1 -2}\\ 3\text{ 1 -1}\end{array}\right]\left[\begin{array}{l}e\\ h\\ k\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$

$\left[\begin{array}{l}1\text{ 0 1}\\ 4\text{ 1 -2}\\ 3\text{ 1 -1}\end{array}\right]\left[\begin{array}{l}f\\ i\\ l\end{array}\right]=\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]$

$\left[\begin{array}{l}1\text{ 0 1}\\ 4\text{ 1 -2}\\ 3\text{ 1 -1}\end{array}\right]\left[\begin{array}{l}g\\ j\\ m\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$

Hence, we get the augmented matrices respectively as follows:

$A|b=\left(\begin{array}{cccc}1& 0& 1& 1\\ 4& 1& -2& 0\\ 3& 1& -1& 0\end{array}\right)$

$B|d=\left(\begin{array}{cccc}1& 0& 1& 0\\ 4& 1& -2& 1\\ 3& 1& -1& 0\end{array}\right)$

$C|v=\left(\begin{array}{cccc}1& 0& 1& 0\\ 4& 1& -2& 0\\ 3& 1& -1& 1\end{array}\right)$

Now, apply Gauss – Jordan method.

Replacing row 2 by (row 2) – 4(row 1) of $A|b$, we get,

 $A_1|b_1=\left(\begin{array}{cccc}1& 0& 1& 1\\ 0& 1& -6& -4\\ 3& 1& -1& 0\end{array}\right)$

Replacing row 3 by (row 3) – 3(row 1) of $A_1|b_1$, we get,

$A_2|b_2=\left(\begin{array}{cccc}1& 0& 1& 1\\ 0& 1& -6& -4\\ 0& 1& -4& -3\end{array}\right)$

Replacing row 3 by (row 3 – row 2) of $A_2|b_2$, we get

$A_3|b_3=\left(\begin{array}{cccc}1& 0& 1& 1\\ 0& 1& -6& -4\\ 0& 0& 2& 1\end{array}\right)$

This gives us,

$\begin{array}{c}e+k=1\\ h-6k=-4\\ k=\frac{1}{2}\end{array}$

Therefore, we get,

$\begin{array}{c}e=\frac{1}{2}\\ h=-1\\ k=\frac{1}{2}\end{array}$

Now, we solve for $f$, $i$ and $l$