Chapter 8.3, Problem 52E

To determine

Tofind:the inverse matrix to solve the system of linear equations.

Answer to Problem 52E

  $x=6\text{ and y=-2}$ .

Explanation of Solution

Given:

  $\begin{array}{l}0.2x-0.6y=2.4\\ -x+1.4y=-8.8\end{array}$

Concept used:

Steps to solve the inverse of matrix by

Step1: calculating the matrix of minors.

Step2:turn that into the matrix of cofactors.

Step3: then ad jugate by transposing the cofactor.

Step4: last multiply by the $\frac{1}{\text{determinant}}$ .

Formula for inverse of matrix A is:

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$

  $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

Where $\text{X=}\left[\begin{array}{c}x\\ y\end{array}\right]$ ,A is left hand side coefficient and B is right hand side coefficient.

Calculation:

Here $\begin{array}{l}0.2x-0.6y=2.4\\ -x+1.4y=-8.8\end{array}$

It can be written on the basis of linear equations in matrix form $\left[\begin{array}{cc}0.2& -0.6\\ -1& 1.4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}2.4\\ -8.8\end{array}\right]$

Let $\text{A=}\left[\begin{array}{cc}0.2& -0.6\\ -1& 1.4\end{array}\right]$ , $\text{X=}\left[\begin{array}{c}x\\ y\end{array}\right]$ and $\text{B=}\left[\begin{array}{c}2.4\\ -8.8\end{array}\right]$ .

Since, $\text{AX=B}$

  ${\text{X=A}}^{\text{-1}}\text{B}$

Finding ${\text{A}}^{-1}$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\text{adjA}}{\left|\text{A}\right|}$

  ${\text{A}}^{\text{-1}}\text{=}\frac{\left[\begin{array}{cc}1.4& 1\\ 0.6& 0.2\end{array}\right]}{-32}=\left[\begin{array}{cc}-4.375& -1.875\\ -3.125& -0.625\end{array}\right]$

Since ${\text{X=A}}^{\text{-1}}\text{B}$

  $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}-4.375& -1.875\\ -3.125& -0.625\end{array}\right]\left[\begin{array}{c}2.4\\ -8.8\end{array}\right]$

  $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-10.5+16.5\\ -7.5+5.5\end{array}\right]\left[\begin{array}{c}2.4\\ -8.8\end{array}\right]$

  $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ -2\end{array}\right]$

Hence comparing matrix of L.H.S and R.H.S

  $x=6\text{ and y=-2}$ .