5. Consider the system below.

{eq}4x - 2y = -12\\ 3x - y = -3 {/eq}

Solve the system by using a matrix equation.

## Question:

5. Consider the system below.

{eq}4x - 2y = -12\\ 3x - y = -3 {/eq}

Solve the system by using a matrix equation.

## Row Operations:

A system of equations may be solved by row operations by writing the system as an augmented matrix. The matrix is transformed into a reduced row echelon matrix to determine the values of each variable from each resulting equation.

## Answer and Explanation: 1

The system of equations may be transformed into an augmented matrix below.

{eq}\left ( \begin{array}{cc|c} 4 & -2 & -12 \\ 3& -1 & -3 \\ \end{array} \right ) {/eq}

Solving the system:

{eq}\left [ R_1/4\Longrightarrow\,R_1 \right ] {/eq}

{eq}\left ( \begin{array}{cc|c} 1 & -1/2 & -3 \\ 3& -1 & -3 \\ \end{array} \right ) {/eq}

{eq}\left [ R_2-3R_1\Longrightarrow\,R_2 \right ] {/eq}

{eq}\left ( \begin{array}{cc|c} 1 & -1/2 & -3 \\ 0& 1/2 & 6 \\ \end{array} \right ) {/eq}

{eq}\left [ R_2/(1/2)\Longrightarrow\,R_2 \right ] {/eq}

{eq}\left ( \begin{array}{cc|c} 1 & -1/2 & -3 \\ 0& 1 & 12 \\ \end{array} \right ) {/eq}

{eq}\left [ R_1-(-1/2)R_2\Longrightarrow\,R_1 \right ] {/eq}

{eq}\left ( \begin{array}{cc|c} 1 & 0 & 3 \\ 0& 1 & 12 \\ \end{array} \right ) {/eq}

Thus, writing the above matrix as a system of equations:

{eq}\begin{align} x&=3\\[0.3cm] y&=12\\[0.3cm] \end{align} {/eq}

The general solution is {eq}\left ( \begin{array}{c} x \\ y\\ \end{array} \right )=\left ( \begin{array}{c} 3\\ 12\\ \end{array} \right ) {/eq}.

#### Learn more about this topic:

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Chapter 10 / Lesson 3Learn about matrix row operations to change matrix using various examples. Understand what a row matrix is. Also, know about the rules for adding matrices.