Linear Algebra Examples

$- x - 4 y = 0$ , $y = - \frac{1}{4} x + 1$

Step 1

Combine $x$ and $\frac{1}{4}$ .

$- x - 4 y = 0, y = - \frac{x}{4} + 1$

Step 2

Add $\frac{x}{4}$ to both sides of the equation.

$- x - 4 y = 0, y + \frac{x}{4} = 1$

Step 3

Write the system of equations in matrix form.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 5

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

$x + 4 y = 0$

$0 = 1$

Step 6

Subtract $4 y$ from both sides of the equation.

$x = - 4 y$

$0 = 1$

Step 7

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

No solution