Linear Algebra Examples

$x + w = 8$ , $2 x = 16$ , $4 x + 8 z = 8$ , $2 y = 4$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 2 & 0 & 0 & 16 \\ 0 & 4 & 0 & 8 & 8 \\ 0 & 0 & 2 & 0 & 4 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ \frac{0}{2} & \frac{2}{2} & \frac{0}{2} & \frac{0}{2} & \frac{16}{2} \\ 0 & 4 & 0 & 8 & 8 \\ 0 & 0 & 2 & 0 & 4 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 4 & 0 & 8 & 8 \\ 0 & 0 & 2 & 0 & 4 \end{array}]$

Step 2.2

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 0 - 4 \cdot 0 & 8 - 4 \cdot 0 & 8 - 4 \cdot 8 \\ 0 & 0 & 2 & 0 & 4 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 0 & 8 & - 24 \\ 0 & 0 & 2 & 0 & 4 \end{array}]$

Step 2.3

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 3$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 2 & 0 & 4 \\ 0 & 0 & 0 & 8 & - 24 \end{array}]$

Step 2.4

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{3}$ by $\frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{0}{2} & \frac{4}{2} \\ 0 & 0 & 0 & 8 & - 24 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 8 & - 24 \end{array}]$

Step 2.5

Multiply each element of $R_{4}$ by $\frac{1}{8}$ to make the entry at $4, 4$ a $1$ .

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Step 2.5.1

Multiply each element of $R_{4}$ by $\frac{1}{8}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 0 & 2 \\ \frac{0}{8} & \frac{0}{8} & \frac{0}{8} & \frac{8}{8} & \frac{- 24}{8} \end{array}]$

Step 2.5.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 0 - 0 & 8 - 8 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 8 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & - 3 \end{array}]$

Step 3

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

$w = 0$

$x = 8$

$y = 2$

$z = - 3$

Step 4

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

$(0, 8, 2, - 3)$

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.

$X = [\begin{array}{c} w \\ x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ 8 \\ 2 \\ - 3 \end{array}]$