Linear Algebra Examples

$1 + 1 - 1 + 1 = 1$ , $3 + 4 + 1 - 3 = - 8$ , $3 + 1 - 1 - 1 = - 15$ , $- 1 + 2 + 1 - 3 = - 2$

Step 1

Simplify.

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Step 1.1

Add $1$ and $1$ .

$2 - 1 + 1 = 1, 3 + 4 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 1.2

Subtract $1$ from $2$ .

$1 + 1 = 1, 3 + 4 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 1.3

Add $1$ and $1$ .

$2 = 1, 3 + 4 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 2

Subtract $2$ from both sides of the equation.

$0 = 1 - 2, 3 + 4 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 3

Subtract $2$ from $1$ .

$0 = - 1, 3 + 4 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 4

Simplify.

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Step 4.1

Add $3$ and $4$ .

$0 = - 1, 7 + 1 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 4.2

Add $7$ and $1$ .

$0 = - 1, 8 - 3 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 4.3

Subtract $3$ from $8$ .

$0 = - 1, 5 = - 8, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 5

Subtract $5$ from both sides of the equation.

$0 = - 1, 0 = - 8 - 5, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 6

Subtract $5$ from $- 8$ .

$0 = - 1, 0 = - 13, 3 + 1 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 7

Simplify.

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Step 7.1

Add $3$ and $1$ .

$0 = - 1, 0 = - 13, 4 - 1 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 7.2

Subtract $1$ from $4$ .

$0 = - 1, 0 = - 13, 3 - 1 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 7.3

Subtract $1$ from $3$ .

$0 = - 1, 0 = - 13, 2 = - 15, - 1 + 2 + 1 - 3 = - 2$

Step 8

Subtract $2$ from both sides of the equation.

$0 = - 1, 0 = - 13, 0 = - 15 - 2, - 1 + 2 + 1 - 3 = - 2$

Step 9

Subtract $2$ from $- 15$ .

$0 = - 1, 0 = - 13, 0 = - 17, - 1 + 2 + 1 - 3 = - 2$

Step 10

Simplify.

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Step 10.1

Add $- 1$ and $2$ .

$0 = - 1, 0 = - 13, 0 = - 17, 1 + 1 - 3 = - 2$

Step 10.2

Add $1$ and $1$ .

$0 = - 1, 0 = - 13, 0 = - 17, 2 - 3 = - 2$

Step 10.3

Subtract $3$ from $2$ .

$0 = - 1, 0 = - 13, 0 = - 17, - 1 = - 2$

Step 11

Add $1$ to both sides of the equation.

$0 = - 1, 0 = - 13, 0 = - 17, 0 = - 2 + 1$

Step 12

Add $- 2$ and $1$ .

$0 = - 1, 0 = - 13, 0 = - 17, 0 = - 1$

Step 13

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 \\ - 13 \\ - 17 \\ - 1 \end{array}]$

Step 14

Find the reduced row echelon form.

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Step 14.1

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

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Step 14.1.1

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

$[\begin{array}{c} - - 1 \\ - 13 \\ - 17 \\ - 1 \end{array}]$

Step 14.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 \\ - 13 \\ - 17 \\ - 1 \end{array}]$

Step 14.2

Perform the row operation $R_{2} = R_{2} + 13 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 14.2.1

Perform the row operation $R_{2} = R_{2} + 13 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 \\ - 13 + 13 \cdot 1 \\ - 17 \\ - 1 \end{array}]$

Step 14.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 \\ 0 \\ - 17 \\ - 1 \end{array}]$

Step 14.3

Perform the row operation $R_{3} = R_{3} + 17 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 14.3.1

Perform the row operation $R_{3} = R_{3} + 17 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 17 + 17 \cdot 1 \\ - 1 \end{array}]$

Step 14.3.2

Simplify $R_{3}$ .

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

Step 14.4

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

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Step 14.4.1

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

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ - 1 + 1 \cdot 1 \end{array}]$

Step 14.4.2

Simplify $R_{4}$ .

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

Step 15

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

$0 = 1$

Step 16

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

No solution