Linear Algebra Examples

$x + y - z = 2$ , $2 x - 3 y + 4 z = - 3$

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 2 \\ 0 & - 5 & 6 & - 7 \end{array}]$

Step 2.2

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & - 1 & 2 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 6 & - \frac{1}{5} \cdot - 7 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 1 & 2 \\ 0 & 1 & - \frac{6}{5} & \frac{7}{5} \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & - 1 + \frac{6}{5} & 2 - \frac{7}{5} \\ 0 & 1 & - \frac{6}{5} & \frac{7}{5} \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{5} & \frac{3}{5} \\ 0 & 1 & - \frac{6}{5} & \frac{7}{5} \end{array}]$

Step 3

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

$x + \frac{z}{5} = \frac{3}{5}$

$y - \frac{6 z}{5} = \frac{7}{5}$

Step 4

Subtract $\frac{z}{5}$ from both sides of the equation.

$x = \frac{3}{5} - \frac{z}{5}$

$y - \frac{6 z}{5} = \frac{7}{5}$

Step 5

Add $\frac{6 z}{5}$ to both sides of the equation.

$y = \frac{7}{5} + \frac{6 z}{5}$

$x = \frac{3}{5} - \frac{z}{5}$

Step 6

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

$(\frac{3}{5} - \frac{z}{5}, \frac{7}{5} + \frac{6 z}{5}, z)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{3}{5} - \frac{z}{5} \\ \frac{7}{5} + \frac{6 z}{5} \\ z \end{array}]$