Q 210 - Intermediate Algebra

Question

Solve Systems of Equations Using Matrices In the following exercises, solve each system of equations using a matrix. 

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

Step-by-Step Solution

Answer

The solution of the system of equations is $(0,2)$

Step 1. Given Information

A system of equations is given  as

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

Step 2. Calculating Matrix

System of equations is given as

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

First convert into argument matrix-

$\left[\begin{array}{rrr}2& 1& 2\\ 1& -1& -2\end{array}\right]$

Then make zeroes in the matrix as-

$$\begin{gathered} \left[\begin{array}{rrr}2& 1& 2\\ 1& -1& -2\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{⟶}\left[\begin{array}{rrr}1& -1& -2\\ 2& 1& 2\end{array}\right] \\ \stackrel{-2R_1+R_2}{⟶}\left[\begin{array}{lll}1& -1& -2\\ (-2\times 1)+2& (-2(-1\left)\right)+1& (-2(-2\left)\right)+2\end{array}\right] \\ \stackrel{-2R_1+R_2}{⟶}\left[\begin{array}{rrr}1& -1& -2\\ 0& 3& 6\end{array}\right] \\ \stackrel{\frac{1}{3}{R}_2}{⟶}\left[\begin{array}{rrr}1& -1& -2\\ 0& 1& 2\end{array}\right] \end{gathered}$$

Step 3. Verification

Verify the equation,

$$\begin{gathered} x-y=-2 \\ 0-2=-2 \\ -2=-2 \end{gathered}$$

LHS is equal to RHS