Linear Algebra Examples

$x + 5 z = 0$ , $y + 2 z = 0$ , $w = 0$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 1 & 0 & 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{3}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 1 & 0 & 5 & 0 \end{array}]$

Step 2.2

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 5 & 0 \\ 0 & 0 & 1 & 2 & 0 \end{array}]$

Step 3

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

$w = 0$

$x + 5 z = 0$

$y + 2 z = 0$

Step 4

Subtract $5 z$ from both sides of the equation.

$x = - 5 z$

$w = 0$

$y + 2 z = 0$

Step 5

Subtract $2 z$ from both sides of the equation.

$y = - 2 z$

$w = 0$

$x = - 5 z$

Step 6

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

$(0, - 5 z, - 2 z, z)$

Step 7

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.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ - 5 z \\ - 2 z \\ z \end{array}]$