Linear Algebra Examples

$z = 3$ , $x = 3 - 5 z$

Step 1

Add $5 z$ to both sides of the equation.

$z = 3, x + 5 z = 3$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 1 & 3 \\ 1 & 5 & 3 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 5 & 3 \\ 0 & 1 & 3 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 - 5 \cdot 0 & 5 - 5 \cdot 1 & 3 - 5 \cdot 3 \\ 0 & 1 & 3 \end{array}]$

Step 3.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 12 \\ 0 & 1 & 3 \end{array}]$

Step 4

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

$x = - 12$

$z = 3$

Step 5

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

$(- 12, 3)$

Step 6

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.

$X = [\begin{array}{c} x \\ z \end{array}] = [\begin{array}{c} - 12 \\ 3 \end{array}]$