Linear Algebra Examples

$x + y + z + w = 15$ , $x - 2 y + 4 z - w = 12$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 1 & 1 & 15 \\ - 1 & 1 & - 2 & 4 & 12 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 1 & 15 \\ - 1 + 1 \cdot 1 & 1 + 1 \cdot 1 & - 2 + 1 \cdot 1 & 4 + 1 \cdot 1 & 12 + 1 \cdot 15 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 15 \\ 0 & 2 & - 1 & 5 & 27 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & 1 & 15 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 1}{2} & \frac{5}{2} & \frac{27}{2} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 15 \\ 0 & 1 & - \frac{1}{2} & \frac{5}{2} & \frac{27}{2} \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 1 + \frac{1}{2} & 1 - \frac{5}{2} & 15 - \frac{27}{2} \\ 0 & 1 & - \frac{1}{2} & \frac{5}{2} & \frac{27}{2} \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} & - \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & - \frac{1}{2} & \frac{5}{2} & \frac{27}{2} \end{array}]$

Step 3

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

$w + \frac{3 y}{2} - \frac{3 z}{2} = \frac{3}{2}$

$x - \frac{y}{2} + \frac{5 z}{2} = \frac{27}{2}$

Step 4

Move all terms not containing $w$ to the right side of the equation.

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Step 4.1

Subtract $\frac{3 y}{2}$ from both sides of the equation.

$w - \frac{3 z}{2} = \frac{3}{2} - \frac{3 y}{2}$

$x - \frac{y}{2} + \frac{5 z}{2} = \frac{27}{2}$

Step 4.2

Add $\frac{3 z}{2}$ to both sides of the equation.

$w = \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}$

$x - \frac{y}{2} + \frac{5 z}{2} = \frac{27}{2}$

$w = \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}$

$x - \frac{y}{2} + \frac{5 z}{2} = \frac{27}{2}$

Step 5

Move all terms not containing $x$ to the right side of the equation.

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Step 5.1

Add $\frac{y}{2}$ to both sides of the equation.

$x + \frac{5 z}{2} = \frac{27}{2} + \frac{y}{2}$

$w = \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}$

Step 5.2

Subtract $\frac{5 z}{2}$ from both sides of the equation.

$x = \frac{27}{2} + \frac{y}{2} - \frac{5 z}{2}$

$w = \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}$

$x = \frac{27}{2} + \frac{y}{2} - \frac{5 z}{2}$

$w = \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}$

Step 6

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

$(\frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2}, \frac{27}{2} + \frac{y}{2} - \frac{5 z}{2}, y, z)$

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.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{3}{2} - \frac{3 y}{2} + \frac{3 z}{2} \\ \frac{27}{2} + \frac{y}{2} - \frac{5 z}{2} \\ y \\ z \end{array}]$