Linear Algebra Examples

$x + y = - 1$ , $x + y = 3$

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{2}$ .

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

Step 2.2

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 3$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 3$ a $1$ .

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{1}$ .

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

Step 3

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

$x + y = 0$

$0 = 1$

Step 4

Subtract $y$ from both sides of the equation.

$x = - y$

$0 = 1$

Step 5

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

No solution