Linear Algebra Examples

$x + y = 30000$ , $0.12 x + 0.15 y = 4140$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 30000 \\ 0.12 & 0.15 & 4140 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation $R_{2} = R_{2} - 0.12 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.1.1

Perform the row operation $R_{2} = R_{2} - 0.12 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 30000 \\ 0.12 - 0.12 \cdot 1 & 0.15 - 0.12 \cdot 1 & 4140 - 0.12 \cdot 30000 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 30000 \\ 0 & 0.03 & 540 \end{array}]$

Step 2.2

Multiply each element of $R_{2}$ by $\frac{1}{0.03}$ to make the entry at $2, 2$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{2}$ by $\frac{1}{0.03}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 30000 \\ \frac{0}{0.03} & \frac{0.03}{0.03} & \frac{540}{0.03} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 30000 \\ 0 & 1 & 18000 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 30000 - 18000 \\ 0 & 1 & 18000 \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 12000 \\ 0 & 1 & 18000 \end{array}]$

Step 3

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

$x = 12000$

$y = 18000$

Step 4

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

$(12000, 18000)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 12000 \\ 18000 \end{array}]$