Linear Algebra Examples

$3 x - x + 2 x = 3$ , $- x + 3 x - 2 x = - 1$ , $2 x + 2 x + 3 x = 14$

Step 1

Simplify.

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Step 1.1

Subtract $x$ from $3 x$ .

$2 x + 2 x = 3, - x + 3 x - 2 x = - 1, 2 x + 2 x + 3 x = 14$

Step 1.2

Add $2 x$ and $2 x$ .

$4 x = 3, - x + 3 x - 2 x = - 1, 2 x + 2 x + 3 x = 14$

Step 2

Simplify.

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Step 2.1

Add $- x$ and $3 x$ .

$4 x = 3, 2 x - 2 x = - 1, 2 x + 2 x + 3 x = 14$

Step 2.2

Subtract $2 x$ from $2 x$ .

$4 x = 3, 0 = - 1, 2 x + 2 x + 3 x = 14$

Step 3

Simplify.

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Step 3.1

Add $2 x$ and $2 x$ .

$4 x = 3, 0 = - 1, 4 x + 3 x = 14$

Step 3.2

Add $4 x$ and $3 x$ .

$4 x = 3, 0 = - 1, 7 x = 14$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & 3 \\ 0 & - 1 \\ 7 & 14 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

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

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Step 5.1.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{3}{4} \\ 0 & - 1 \\ 7 & 14 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} \\ 0 & - 1 \\ 7 & 14 \end{array}]$

Step 5.2

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 5.2.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} \\ 0 & - 1 \\ 7 - 7 \cdot 1 & 14 - 7 (\frac{3}{4}) \end{array}]$

Step 5.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} \\ 0 & - 1 \\ 0 & \frac{35}{4} \end{array}]$

Step 5.3

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

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Step 5.3.1

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

$[\begin{array}{c} 1 & \frac{3}{4} \\ - 0 & - - 1 \\ 0 & \frac{35}{4} \end{array}]$

Step 5.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} \\ 0 & 1 \\ 0 & \frac{35}{4} \end{array}]$

Step 5.4

Perform the row operation $R_{3} = R_{3} - \frac{35}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.4.1

Perform the row operation $R_{3} = R_{3} - \frac{35}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} \\ 0 & 1 \\ 0 - \frac{35}{4} \cdot 0 & \frac{35}{4} - \frac{35}{4} \cdot 1 \end{array}]$

Step 5.4.2

Simplify $R_{3}$ .

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

Step 5.5

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 5.5.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 5.5.2

Simplify $R_{1}$ .

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

Step 6

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

$x = 0$

$0 = 1$

Step 7

Since $0 \neq 1$ , there are no solutions.

No solution