Linear Algebra Examples

$- 4 + y = 4$ , $4 x - y = 4$

Step 1

Add $4$ to both sides of the equation.

$y = 4 + 4, 4 x - y = 4$

Step 2

Add $4$ and $4$ .

$y = 8, 4 x - y = 4$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 1 & 8 \\ 4 & - 1 & 4 \end{array}]$

Step 4

Find the reduced row echelon form.

Tap for more steps...

Step 4.1

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

$[\begin{array}{c} 4 & - 1 & 4 \\ 0 & 1 & 8 \end{array}]$

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{4}{4} \\ 0 & 1 & 8 \end{array}]$

Step 4.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 1 \\ 0 & 1 & 8 \end{array}]$

Step 4.3

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 4.3.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & 1 + \frac{1}{4} \cdot 8 \\ 0 & 1 & 8 \end{array}]$

Step 4.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 3 \\ 0 & 1 & 8 \end{array}]$

Step 5

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

$x = 3$

$y = 8$

Step 6

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

$(3, 8)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 \\ 8 \end{array}]$