Linear Algebra Examples

$x = - 3 y$ , $7 x - 2 y = - 69$

Step 1

Add $3 y$ to both sides of the equation.

$x + 3 y = 0, 7 x - 2 y = - 69$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & 0 \\ 7 & - 2 & - 69 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 0 \\ 7 - 7 \cdot 1 & - 2 - 7 \cdot 3 & - 69 - 7 \cdot 0 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 0 \\ 0 & - 23 & - 69 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $- \frac{1}{23}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{23}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 0 \\ - \frac{1}{23} \cdot 0 & - \frac{1}{23} \cdot - 23 & - \frac{1}{23} \cdot - 69 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 0 \\ 0 & 1 & 3 \end{array}]$

Step 3.3

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 3 \\ 0 & 1 & 3 \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 9 \\ 0 & 1 & 3 \end{array}]$

Step 4

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

$x = - 9$

$y = 3$

Step 5

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

$(- 9, 3)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 9 \\ 3 \end{array}]$