Linear Algebra Examples

$4 x - 2 y = - 2$ , $- 5 x + 6 y = - 22$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & - 2 & - 2 \\ - 5 & 6 & - 22 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 2}{4} & \frac{- 2}{4} \\ - 5 & 6 & - 22 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{1}{2} \\ - 5 & 6 & - 22 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{1}{2} \\ - 5 + 5 \cdot 1 & 6 + 5 (- \frac{1}{2}) & - 22 + 5 (- \frac{1}{2}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & \frac{7}{2} & - \frac{49}{2} \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{2}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{2}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{1}{2} \\ \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} (- \frac{49}{2}) \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & - 7 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - \frac{1}{2} + \frac{1}{2} \cdot - 7 \\ 0 & 1 & - 7 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 4 \\ 0 & 1 & - 7 \end{array}]$

Step 3

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

$x = - 4$

$y = - 7$

Step 4

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

$(- 4, - 7)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 4 \\ - 7 \end{array}]$