Linear Algebra Examples

$3 x = - 3 (3 y - 4)$ , $9 x + 27 = 36$

Step 1

Simplify.

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Step 1.1

Apply the distributive property.

$3 x = - 3 (3 y) - 3 \cdot - 4, 9 x + 27 = 36$

Step 1.2

Multiply $3$ by $- 3$ .

$3 x = - 9 y - 3 \cdot - 4, 9 x + 27 = 36$

Step 1.3

Multiply $- 3$ by $- 4$ .

$3 x = - 9 y + 12, 9 x + 27 = 36$

Step 2

Add $9 y$ to both sides of the equation.

$3 x + 9 y = 12, 9 x + 27 = 36$

Step 3

Subtract $27$ from both sides of the equation.

$3 x + 9 y = 12, 9 x = 36 - 27$

Step 4

Subtract $27$ from $36$ .

$3 x + 9 y = 12, 9 x = 9$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 3 & 9 & 12 \\ 9 & 0 & 9 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

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

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Step 6.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{9}{3} & \frac{12}{3} \\ 9 & 0 & 9 \end{array}]$

Step 6.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 4 \\ 9 & 0 & 9 \end{array}]$

Step 6.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 6.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 4 \\ 9 - 9 \cdot 1 & 0 - 9 \cdot 3 & 9 - 9 \cdot 4 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 4 \\ 0 & - 27 & - 27 \end{array}]$

Step 6.3

Multiply each element of $R_{2}$ by $- \frac{1}{27}$ to make the entry at $2, 2$ a $1$ .

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Step 6.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{27}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 4 \\ - \frac{1}{27} \cdot 0 & - \frac{1}{27} \cdot - 27 & - \frac{1}{27} \cdot - 27 \end{array}]$

Step 6.3.2

Simplify $R_{2}$ .

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

Step 6.4

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 6.4.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 4 - 3 \cdot 1 \\ 0 & 1 & 1 \end{array}]$

Step 6.4.2

Simplify $R_{1}$ .

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

Step 7

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

$x = 1$

$y = 1$

Step 8

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

$(1, 1)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ 1 \end{array}]$