Linear Algebra Examples

$x - y = 7$ , $2 x - 2 y = k$

Step 1

Subtract $k$ from both sides of the equation.

$x - y = 7, 2 x - 2 y - k = 0$

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{1}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 14 \\ 0 & 1 & - 1 & 7 \end{array}]$

Step 4

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

$k = 14$

$x - y = 7$

Step 5

Add $y$ to both sides of the equation.

$x = 7 + y$

$k = 14$

Step 6

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

$(14, 7 + y, y)$

Step 7

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.

$X = [\begin{array}{c} k \\ x \\ y \end{array}] = [\begin{array}{c} 14 \\ 7 + y \\ y \end{array}]$