Linear Algebra Examples

$d = 5 (2 w + 1120)$ , $d = 7 (1120)$

Step 1

Simplify.

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Step 1.1

Apply the distributive property.

$d = 5 (2 w) + 5 \cdot 1120, d = 7 (1120)$

Step 1.2

Multiply $2$ by $5$ .

$d = 10 w + 5 \cdot 1120, d = 7 (1120)$

Step 1.3

Multiply $5$ by $1120$ .

$d = 10 w + 5600, d = 7 (1120)$

Step 2

Subtract $10 w$ from both sides of the equation.

$d - 10 w = 5600, d = 7 (1120)$

Step 3

Multiply $7$ by $1120$ .

$d - 10 w = 5600, d = 7840$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 10 & 5600 \\ 1 & 0 & 7840 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} 1 & - 10 & 5600 \\ 1 - 1 & 0 + 10 & 7840 - 5600 \end{array}]$

Step 5.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 10 & 5600 \\ 0 & 10 & 2240 \end{array}]$

Step 5.2

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

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Step 5.2.1

Multiply each element of $R_{2}$ by $\frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 10 & 5600 \\ \frac{0}{10} & \frac{10}{10} & \frac{2240}{10} \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 10 & 5600 \\ 0 & 1 & 224 \end{array}]$

Step 5.3

Perform the row operation $R_{1} = R_{1} + 10 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 5.3.1

Perform the row operation $R_{1} = R_{1} + 10 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 10 \cdot 0 & - 10 + 10 \cdot 1 & 5600 + 10 \cdot 224 \\ 0 & 1 & 224 \end{array}]$

Step 5.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 7840 \\ 0 & 1 & 224 \end{array}]$

Step 6

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

$d = 7840$

$w = 224$

Step 7

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

$(7840, 224)$

Step 8

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.

$X = [\begin{array}{c} d \\ w \end{array}] = [\begin{array}{c} 7840 \\ 224 \end{array}]$