Linear Algebra Examples

$x + y = k$ , $x - y = k$

Step 1

Subtract $k$ from both sides of the equation.

$x + y - k = 0, x - y = k$

Step 2

Subtract $k$ from both sides of the equation.

$x + y - k = 0, x - y - k = 0$

Step 3

Write the system of equations in matrix form.

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

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 $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify $R_{2}$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify $R_{1}$ .

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

Step 5

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

$k - x = 0$

$y = 0$

Step 6

Add $x$ to both sides of the equation.

$k = x$

$y = 0$

Step 7

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

$(x, x, 0)$

Step 8

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.

$X = [\begin{array}{c} k \\ x \\ y \end{array}] = [\begin{array}{c} x \\ x \\ 0 \end{array}]$