Linear Algebra Examples

$y = m x + b$ , $- 1 = m x + 2$

Step 1

Move all terms containing variables to the left side of the equation.

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Step 1.1

Subtract $m x$ from both sides of the equation.

$y - m x = b, - 1 = m x + 2$

Step 1.2

Subtract $b$ from both sides of the equation.

$y - m x - b = 0, - 1 = m x + 2$

Step 2

Add $1$ to both sides of the equation.

$y - m x - b = 0, 0 = m x + 2 + 1$

Step 3

Subtract $m x$ from both sides of the equation.

$y - m x - b = 0, - m x = 2 + 1$

Step 4

Write the system of equations in matrix form.

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

Step 5

Find the reduced row echelon form.

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Step 5.1

Rewrite $- 1 x$ as $- x$ .

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

Step 5.2

Rewrite $- 1 x$ as $- x$ .

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

Step 5.3

Add $2$ and $1$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify $R_{1}$ .

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

Simplify $R_{2}$ .

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

Step 5.6

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

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Step 5.6.1

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

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

Step 5.6.2

Simplify $R_{1}$ .

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

Step 6

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

$b - y = 3$

$m = - \frac{3}{x}$

Step 7

Add $y$ to both sides of the equation.

$b = 3 + y$

$m = - \frac{3}{x}$

Step 8

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

$(3 + y, - \frac{3}{x}, x, y)$

Step 9

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.

$X = [\begin{array}{c} b \\ m \\ x \\ y \end{array}] = [\begin{array}{c} 3 + y \\ - \frac{3}{x} \\ x \\ y \end{array}]$