Linear Algebra Examples

$y = x - 4$ , $x + 3 y = 12$

Step 1

Subtract $x$ from both sides of the equation.

$y - x = - 4, x + 3 y = 12$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 1 & - 4 \\ 1 & 3 & 12 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - - 4 \\ 1 & 3 & 12 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 4 \\ 1 & 3 & 12 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - 1 & 4 \\ 1 - 1 & 3 + 1 & 12 - 4 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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

Step 3.3.2

Simplify $R_{2}$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 6$

$y = 2$

Step 5

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

$(6, 2)$

Step 6

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 6 \\ 2 \end{array}]$