Chapter 11, Problem 20RE

To determine

To find: The value of $x$ and $y$ .

Answer to Problem 20RE

The value of $x$ is $\frac{2}{5}$

The value of $y$ is $\frac{1}{10}$

Explanation of Solution

Given information:

  $\begin{array}{l}3x-2y=1\\ 10x+10y=5\end{array}$

Calculation:

In mathematics

  $\begin{array}{l}\left[\begin{array}{l}3-2\\ 1010\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}1\\ 5\end{array}\right]\\ \Rightarrow AX=B\\ \Rightarrow X=A^{-1}B---\left(i\right)\end{array}$

Now,

  $\begin{array}{l}{A}^{-1}=\frac{adj\left(A\right)}{\left|A\right|}\left[where,\left|A\right|\ne 0\right]\\ \Rightarrow \left|A\right|=\left(3\times 10\right)-\left(-2\times 10\right)\\ \Rightarrow \left|A\right|=30+20\\ \Rightarrow \left|A\right|=50\end{array}$

  $\begin{array}{l}adj\left(A\right)=\left[\begin{array}{l}102\\ -10\text{ 3}\end{array}\right]\\ \Rightarrow A^{-1}=\frac{adj\left(A\right)}{\left|A\right|}\\ \Rightarrow A^{-1}=\frac{1}{50}\left[\begin{array}{l}102\\ -10\text{ 3}\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}102\\ -10\text{ 3}\end{array}\right]\left[\begin{array}{l}1\\ 5\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}10\times 12\times 5\\ -10\times 1\text{ 3}\times \text{5}\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}20\\ 5\end{array}\right]\\ \Rightarrow \left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{2}{5}\\ \frac{1}{10}\end{array}\right]\end{array}$

  $\begin{array}{l}\therefore x=\frac{2}{5}\\ y=\frac{1}{10}\end{array}$

Hence,

The value of $x$ is $\frac{2}{5}$

The value of $y$ is $\frac{1}{10}$