Q 212 - Intermediate Algebra

Question

Solve Systems of Equations Using Matrices

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

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

Step-by-Step Solution

Answer

The solution of the system is $(-2,-2)$

Step 1. Given

A system of linear equations is given as-

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

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

Step 3. Calculation

A system of linear equations is given as

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

Change it into an argument matrix

$\left[\begin{array}{rrr}-1& 2& -2\\ 1& 1& -4\end{array}\right]$

Apply transformations as-

$$\begin{gathered} \stackrel{R_1+R_2}{⟶}\left[\begin{array}{lll}-1& 2& -2\\ -1+1& 2+1& -2-4\end{array}\right] \\ \stackrel{R_1+{\mathit{R}}_2}{⟶}\left[\begin{array}{rrr}-1& 2& -2\\ 0& 3& -6\end{array}\right] \\ \stackrel{\frac{1}{3}{R}_2}{⟶}\left[\begin{array}{lll}-1& 2& -2\\ 0& 1& -2\end{array}\right] \end{gathered}$$

Step 4. Further Calculation

Corresponds equations are-

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

Use substitution.

$$\begin{gathered} -x+2y=-2 \\ y=-2 \\ -x+2(-2)=-2 \\ -x-4=-2 \\ x+4=2 \\ x=2-4 \\ x=-2 \end{gathered}$$

Step 5. Verification

Verify the equation,

$$\begin{gathered} -x+2 y=-2 \\ 2+2(-2)=-2 \\ 2-4=-2 \\ -2=-2 \end{gathered}$$

LHS is equal to RHS