Linear Algebra Examples

$x + 2 y + 3 z = 90$ , $3 y + 4 z = 36$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 2 & 3 & 90 \\ 0 & 3 & 4 & 36 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 2 & 3 & 90 \\ \frac{0}{3} & \frac{3}{3} & \frac{4}{3} & \frac{36}{3} \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 90 \\ 0 & 1 & \frac{4}{3} & 12 \end{array}]$

Step 2.2

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 2.2.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 3 - 2 (\frac{4}{3}) & 90 - 2 \cdot 12 \\ 0 & 1 & \frac{4}{3} & 12 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{3} & 66 \\ 0 & 1 & \frac{4}{3} & 12 \end{array}]$

Step 3

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

$x + \frac{z}{3} = 66$

$y + \frac{4 z}{3} = 12$

Step 4

Subtract $\frac{z}{3}$ from both sides of the equation.

$x = 66 - \frac{z}{3}$

$y + \frac{4 z}{3} = 12$

Step 5

Subtract $\frac{4 z}{3}$ from both sides of the equation.

$y = 12 - \frac{4 z}{3}$

$x = 66 - \frac{z}{3}$

Step 6

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

$(66 - \frac{z}{3}, 12 - \frac{4 z}{3}, z)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 66 - \frac{z}{3} \\ 12 - \frac{4 z}{3} \\ z \end{array}]$