Linear Algebra Examples

$x + y = 65$ , $30 x + 40 y = 2200$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 65 \\ 30 & 40 & 2200 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation $R_{2} = R_{2} - 30 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.1.1

Perform the row operation $R_{2} = R_{2} - 30 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 65 \\ 30 - 30 \cdot 1 & 40 - 30 \cdot 1 & 2200 - 30 \cdot 65 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 65 \\ 0 & 10 & 250 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 1 & 65 \\ \frac{0}{10} & \frac{10}{10} & \frac{250}{10} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 65 \\ 0 & 1 & 25 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 65 - 25 \\ 0 & 1 & 25 \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 40 \\ 0 & 1 & 25 \end{array}]$

Step 3

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

$x = 40$

$y = 25$

Step 4

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

$(40, 25)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 40 \\ 25 \end{array}]$