Linear Algebra Examples

$A = [\begin{array}{c} 3 & 1 & 2 \end{array}]$ , $x = [\begin{array}{c} x & 3 y & z \end{array}]$

Step 1

Write as a linear system of equations.

$3 = x$

$1 = 3 y$

$2 = z$

Step 2

Solve the system of equations.

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Step 2.1

Move variables to the left and constant terms to the right.

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Step 2.1.1

Subtract $x$ from both sides of the equation.

$3 - x = 0$

$1 = 3 y$

$2 = z$

Step 2.1.2

Subtract $3$ from both sides of the equation.

$- x = - 3$

$1 = 3 y$

$2 = z$

Step 2.1.3

Subtract $3 y$ from both sides of the equation.

$- x = - 3$

$1 - 3 y = 0$

$2 = z$

Step 2.1.4

Subtract $1$ from both sides of the equation.

$- x = - 3$

$- 3 y = - 1$

$2 = z$

Step 2.1.5

Subtract $z$ from both sides of the equation.

$- x = - 3$

$- 3 y = - 1$

$2 - z = 0$

Step 2.1.6

Subtract $2$ from both sides of the equation.

$- x = - 3$

$- 3 y = - 1$

$- z = - 2$

$- x = - 3$

$- 3 y = - 1$

$- z = - 2$

Step 2.2

Write the system as a matrix.

$[\begin{array}{c} - 1 & 0 & 0 & - 3 \\ 0 & - 3 & 0 & - 1 \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.3

Find the reduced row echelon form.

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Step 2.3.1

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

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Step 2.3.1.1

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

$[\begin{array}{c} - - 1 & - 0 & - 0 & - - 3 \\ 0 & - 3 & 0 & - 1 \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ 0 & - 3 & 0 & - 1 \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.3.2

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

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Step 2.3.2.1

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

$[\begin{array}{c} 1 & 0 & 0 & 3 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 1 \\ 0 & 0 & - 1 & - 2 \end{array}]$

Step 2.3.2.2

Simplify $R_{2}$ .

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

Step 2.3.3

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

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Step 2.3.3.1

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

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

Step 2.3.3.2

Simplify $R_{3}$ .

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

Step 2.4

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

$x = 3$

$y = \frac{1}{3}$

$z = 2$

Step 2.5

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 3 \\ \frac{1}{3} \\ 2 \end{array}]$

Step 2.6

Write as a solution set.

${[\begin{array}{c} 3 \\ \frac{1}{3} \\ 2 \end{array}]}$

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