Linear Algebra Examples

$- 18 x + 6 y = 24$ , $3 x - y = - 2$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} - 18 & 6 & 24 \\ 3 & - 1 & - 2 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{18}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{18}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{18} \cdot - 18 & - \frac{1}{18} \cdot 6 & - \frac{1}{18} \cdot 24 \\ 3 & - 1 & - 2 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{4}{3} \\ 3 & - 1 & - 2 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{4}{3} \\ 3 - 3 \cdot 1 & - 1 - 3 (- \frac{1}{3}) & - 2 - 3 (- \frac{4}{3}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 0 & 2 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{4}{3} \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 0 & 1 \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{1}{3} + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

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

Step 3

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

$x - \frac{y}{3} = 0$

$0 = 1$

Step 4

Add $\frac{y}{3}$ to both sides of the equation.

$x = \frac{y}{3}$

$0 = 1$

Step 5

Since $0 \neq 1$ , there are no solutions.

No solution