Q 213 - Intermediate Algebra

Question

Solve Systems of Equations Using Matrices

In the following exercises, solve each system of equations using a matrix.

$\left\{\begin{array}{l}-2x+3y=3\\ x+3y=12\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations is $(3,3)$

Step 1. Given

A system of linear equations is given as-

$\left\{\begin{array}{l}-2x+3y=3\\ x+3y=12\end{array}\right.$

Step 2. Concept Used

First, we will convert the given system of linear equations into its argument matrix. After that we will apply row transformation accordingly. 

And we will to make zeroes in the matrix at least one through which we can find the value of one of the variable. And then we will find the value of other variable by using value of first one.

Step 3. Calculation

A system of linear equations is given as-

$\left\{\begin{array}{l}-2x+3y=3\\ x+3y=12\end{array}\right.$

Correspond argument matrix is-

$\left[\begin{array}{ccc}-2& 3& 3\\ 1& 3& 12\end{array}\right]$

Apply row transformations as-

$$\begin{gathered} \left[\begin{array}{rrc}-2& 3& 3\\ 1& 3& 12\end{array}\right]\stackrel{R_{\leftrightarrow }\leftrightarrow R_2}{⟶}\left[\begin{array}{rrc}1& 3& 12\\ -2& 3& 3\end{array}\right] \\ \stackrel{2R_1+R_2}{⟶}\left[\begin{array}{lll}1& 3& 12\\ 0& 9& 27\end{array}\right] \\ \stackrel{\frac{1}{9}{R}_2}{⟶}\left[\begin{array}{lll}1& 3& 12\\ 0& 1& 3\end{array}\right] \end{gathered}$$

Step 4. Further Calculation

Accordingly, the equations are-

$$\begin{gathered} x+3 y=12 \\ y=3 \end{gathered}$$

Use substitution,

$$\begin{gathered} x+3 y=12 \\ x+3\left(3\right)=12 \\ x+9=12 \\ x=12-9 \\ x=3 \end{gathered}$$

Step 5. Verification

Verify the equation,

$$\begin{gathered} -2 x+3 y=3 \\ -2\left(3\right)+3\left(3\right)=3 \\ -6+9=3 \\ 3=3 \end{gathered}$$