Algebra Examples

$4 x - y = - 4$ , $3 x - 3 y = - 6$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & - 1 & - 4 \\ 3 & - 3 & - 6 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{- 4}{4} \\ 3 & - 3 & - 6 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 3 & - 3 & - 6 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 3 - 3 \cdot 1 & - 3 - 3 (- \frac{1}{4}) & - 6 - 3 \cdot - 1 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 0 & - \frac{9}{4} & - 3 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $- \frac{4}{9}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $- \frac{4}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ - \frac{4}{9} \cdot 0 & - \frac{4}{9} (- \frac{9}{4}) & - \frac{4}{9} \cdot - 3 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 0 & 1 & \frac{4}{3} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & - 1 + \frac{1}{4} \cdot \frac{4}{3} \\ 0 & 1 & \frac{4}{3} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{2}{3} \\ 0 & 1 & \frac{4}{3} \end{array}]$

Step 3

Use the result matrix to declare the final solutions to the system of equations.

$x = - \frac{2}{3}$

$y = \frac{4}{3}$

Step 4

The solution is the set of ordered pairs that makes the system true.

$(- \frac{2}{3}, \frac{4}{3})$

Step 5

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{2}{3} \\ \frac{4}{3} \end{array}]$

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