Chapter E, Problem 1ES

To determine

To solve:the givenset of linearequationfor value of variable using matrices and mathematical methodologies.

Answer to Problem 1ES

The value of variable is $x=2andy=-1$ .

Explanation of Solution

Given information:

The given is below set of linear equation:

  $\left\{\begin{array}{l}x+y=1\\ x-2y=4\end{array}\right\}$

Calculation:

The given is below set of linear equation:

  $\left\{\begin{array}{l}x+y=1\\ x-2y=4\end{array}\right\}$

The given equation is re-written below in matrices form:

  $\begin{array}{l}\left\{\begin{array}{l}x+y=1\\ x-2y=4\end{array}\right\}\\ \left(\begin{array}{cc}1& 1\\ 1& -2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}1\\ 4\end{array}\right)\end{array}$

This matrix representation is in form of:

  $AX=B$

Here matrix A represents coefficients of linear equation, matrix B represents the resultant of two equation and matrix $X$ represents variables of linear equations. Applying matrix properties as below:

  $\begin{array}{l}AX=B\\ A^{-1}AX=A^{-1}B\left[\text{multiply both side by}{A}^{-1}\right]\\ IX=A^{-1}B\left[I\text{is identity matrix and }IX=X\right]\\ X=A^{-1}B\end{array}$

So, the value of variables of linear equation is equal to product of inverse of matrix A and matrix B.

Now it is known that inverse of matrix is equal to adjoint of A divided by determinant of matrix A, as below:

  $A^{-1}=\frac{1}{\left|A\right|}adj(A)$

Put value in above formula to find value of A inverse for given linear equation:

  $\begin{array}{l}{A}^{-1}=\frac{1}{(1)(-2)-(1)(1)}\left(\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right)\\ A^{-1}=\frac{1}{-2-1}\left(\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right)\\ A^{-1}=\frac{1}{-3}\left(\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right)\end{array}$

Put value A inverse and B in formula for variables, as below:

  $\begin{array}{l}X=A^{-1}B\\ \left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{-3}\left(\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}1\\ 4\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{-3}\left(\begin{array}{c}-6\\ 3\end{array}\right)\left[\text{multiply the matrices first}\right]\\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}2\\ -1\end{array}\right)\end{array}$

Hence, the value of variable is $x=2andy=-1$ .