Chapter 4.6, Problem 30PPS

To determine

To solve: the matrix equations.

Answer to Problem 30PPS

The solution is $\left(\frac{3}{4},\frac{3}{4}\right)$ .

Explanation of Solution

Given information:

  $\begin{array}{l}3x+y=3\\ 5x+3y=6\end{array}$ .

Formula used:

Inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ is, where $ad-bc\ne 0$ .

The value of a second order determinant is the difference of the products of the products of the two diagonals.

Steps to solve the equations are:-

Step1. Find the inverse of the coefficient matrix.

Step 2. Multiple each side of the matrix equation by the inverse matrix.

Calculation:

Consider,

  $\begin{array}{l}3x+y=3\\ 5x+3y=6\end{array}$

The matrix equation is

  $\left[\begin{array}{cc}3& 1\\ 5& 3\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 6\end{array}\right]$ .

Step 1.

  $=A^{-1}=\frac{1}{(3)(3)-(1)(5)}\left[\begin{array}{cc}3& -1\\ -5& 3\end{array}\right]$

  $\begin{array}{l}=\frac{1}{9-5}\left[\begin{array}{cc}3& -1\\ -5& 3\end{array}\right]\\ =\frac{1}{4}\left[\begin{array}{cc}3& -1\\ -5& 3\end{array}\right]\end{array}$

Step2.

  $\begin{array}{l}=\frac{1}{4}\left[\begin{array}{cc}3& -1\\ -5& 3\end{array}\right]\left[\begin{array}{cc}3& 1\\ 5& 3\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}3& -1\\ -5& 3\end{array}\right]\left[\begin{array}{c}3\\ 6\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}(3\times 3)+(-1\times 6)\\ (-5\times 3)+(3\times 6)\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}9-6\\ -15+18\end{array}\right]\\ =\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}3\\ 3\end{array}\right]\end{array}$

Or

  $=\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{3}{4}\\ \frac{3}{4}\end{array}\right]$

Thus, the solution is $\left(\frac{3}{4},\frac{3}{4}\right)$ .