Linear Algebra Examples

$y = \frac{1}{2} x - 3$ , $2 x + y = 2$

Step 1

Combine $\frac{1}{2}$ and $x$ .

$y = \frac{x}{2} - 3, 2 x + y = 2$

Step 2

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

$y - \frac{x}{2} = - 3, 2 x + y = 2$

Step 3

Write the system of equations in matrix form.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

$[\begin{array}{c} - 2 (- \frac{1}{2}) & - 2 \cdot 1 & - 2 \cdot - 3 \\ 2 & 1 & 2 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & 6 \\ 2 & 1 & 2 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & - 2 & 6 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 2 & 2 - 2 \cdot 6 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 6 \\ 0 & 5 & - 10 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2 & 6 \\ \frac{0}{5} & \frac{5}{5} & \frac{- 10}{5} \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 6 \\ 0 & 1 & - 2 \end{array}]$

Step 4.4

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

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Step 4.4.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 6 + 2 \cdot - 2 \\ 0 & 1 & - 2 \end{array}]$

Step 4.4.2

Simplify $R_{1}$ .

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

Step 5

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

$x = 2$

$y = - 2$

Step 6

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

$(2, - 2)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ - 2 \end{array}]$