Chapter 8.3, Problem 40E

To determine

To calculate: The solution set of the system of equations,

  $\{\begin{array}{c}x-2y=0\\ 2x-3y=3\end{array}$

With help of the inverse of coefficient matrix.

Answer to Problem 40E

The solution set of the system of equations is $\left(6,3\right)$ .

Explanation of Solution

Given information:

The system of equations is provided below,

  $\{\begin{array}{c}x-2y=0\\ 2x-3y=3\end{array}$

Formula used:

The solution set of the system of equations with same number of equations and variables is given by, $X=P^{-1}Q$ , where, X represents the variable matrix, P represent the coefficient matrix corresponding to variables, and Q represents the matrix of constants given on right hand of the equation.

For a matrix A , the inverse of a $2\times 2$ is given by the formula $A{A}^{-1}=I_2$ , where $I_2$ is the identity matrix, we use the Gauss-Jordan elimination to find the inverse.

Calculation:

Consider the system of equations as,

  $\{\begin{array}{c}x-2y=0\\ 2x-3y=3\end{array}$

Now, in the above system of equation we have two equations and two variables x and y.

Construct a coefficient matrix P ,

  $P=\left[\begin{array}{ll}1\hfill & -2\hfill \\ 2\hfill & -3\hfill \end{array}\right]$

Secondly, the variable matrix X is given as,

  $X=\left[\begin{array}{c}x\\ y\end{array}\right]$

Thirdly, matrix of constants Q which is given on right hand of the equation is,

  $Q=\left[\begin{array}{c}0\\ 3\end{array}\right]$

Therefore, write the system as,

  $\begin{array}{c}PX=Q\\ \left[\begin{array}{ll}1\hfill & -2\hfill \\ 2\hfill & -3\hfill \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 3\end{array}\right]\end{array}$

First, find the inverse of the matrix, $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ .

Let the inverse of the matrix be $P^{-1}=\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]$ .

Use the relation, $P{P}^{-1}=I_2$ , to get,

  $\left[\begin{array}{ll}1\hfill & -2\hfill \\ 2\hfill & -3\hfill \end{array}\right]\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, multiply the above two matrices, to get,

  $\left[\begin{array}{ll}a-2b\hfill & c-2d\hfill \\ 2a-3b\hfill & 2c-3d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, equate the corresponding elements of the two matrices. First write the system of equations when equated to the first column,

  $\begin{array}{c}a-2b=1\\ 2a-3b=0\end{array}$

Now, write the system of equations when equated to the second column,

  $\begin{array}{c}c-2d=0\\ 2c-3d=1\end{array}$

So, get the two augmented matrices which are as follows:

  $\left[\begin{array}{lll}1\hfill & -2\hfill & 1\hfill \\ 2\hfill & -3\hfill & 0\hfill \end{array}\right]$ and $\left[\begin{array}{lll}1\hfill & -2\hfill & 0\hfill \\ 2\hfill & -3\hfill & 1\hfill \end{array}\right]$

Combing the above two matrices such that the identity matrix is on the right hand side, to get,

  $\left[\begin{array}{llll}1\hfill & -2\hfill & 1\hfill & 0\hfill \\ 2\hfill & -3\hfill & 0\hfill & 1\hfill \end{array}\right]$

Now, apply the Gauss-Jordan elimination to find the inverse of $\left[\begin{array}{llll}1\hfill & -2\hfill & 1\hfill & 0\hfill \\ 2\hfill & -3\hfill & 0\hfill & 1\hfill \end{array}\right]$ ,

First to make the entry $a_{11}$ as 1, no need to perform row transformation as it is already one.

Next, to make the entry $a_{21}$ as 0, perform the row transformation as $R_2\to R_2-2R_1$ .

  $\left[\begin{array}{llll}1\hfill & -2\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & -2\hfill & 1\hfill \end{array}\right]$

Next, to make the entry $a_{22}$ as 1, no need to perform row transformation as it is already one.

Next, to make the entry $a_{12}$ as 0, perform the row transformation as $R_1\to R_1+2R_2$ .

  $\left[\begin{array}{llll}1\hfill & 0\hfill & -3\hfill & 2\hfill \\ 0\hfill & 1\hfill & -2\hfill & 1\hfill \end{array}\right]$

Therefore, the inverse of the matrix $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ is $P^{{}^{-1}}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]$ .

Now, substituting the values of $P^{{}^{-1}}$ , Q , and X in the equation $X=P^{-1}Q$ ,

  $\begin{array}{c}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\left[\begin{array}{c}0\\ 3\end{array}\right]\\ =\left[\begin{array}{c}\left(-3\right)\left(0\right)+\left(2\right)\left(3\right)\\ \left(-2\right)\left(0\right)+\left(1\right)\left(3\right)\end{array}\right]\\ =\left[\begin{array}{c}0+6\\ 0+3\end{array}\right]\\ =\left[\begin{array}{c}6\\ 3\end{array}\right]\end{array}$

Therefore, the value of x is 6 and y is $3$ for the system of equations.

Thus, the solution set of the system of equations is $\left(6,3\right)$ .