Linear Algebra Examples

$2 x - 5 = 20$ , $y = x - 7$

Step 1

Add $5$ to both sides of the equation.

$2 x = 20 + 5, y = x - 7$

Step 2

Add $20$ and $5$ .

$2 x = 25, y = x - 7$

Step 3

Subtract $x$ from both sides of the equation.

$2 x = 25, y - x = - 7$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & 0 & 25 \\ - 1 & 1 & - 7 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{0}{2} & \frac{25}{2} \\ - 1 & 1 & - 7 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{25}{2} \\ - 1 & 1 & - 7 \end{array}]$

Step 5.2

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

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Step 5.2.1

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

$[\begin{array}{c} 1 & 0 & \frac{25}{2} \\ - 1 + 1 \cdot 1 & 1 + 0 & - 7 + \frac{25}{2} \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{25}{2} \\ 0 & 1 & \frac{11}{2} \end{array}]$

Step 6

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

$x = \frac{25}{2}$

$y = \frac{11}{2}$

Step 7

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

$(\frac{25}{2}, \frac{11}{2})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{25}{2} \\ \frac{11}{2} \end{array}]$