Linear Algebra Examples

$A = [\begin{array}{c} - 1 \\ 15 \\ 2 \end{array}]$ , $x = [\begin{array}{c} 16 \\ - 2 \\ 3 \end{array}]$

Move all terms not containing a variable to the right side of the equation.

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Add $1$ to both sides of the equation.

$0 = 16 + 1$

$15 = - 2$

$2 = 3$

Add $16$ and $1$ .

$0 = 17$

$15 = - 2$

$2 = 3$

$0 = 17$

$15 = - 2$

$2 = 3$

Move all terms not containing a variable to the right side of the equation.

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Subtract $15$ from both sides of the equation.

$0 = 17$

$0 = - 2 - 15$

$0 = 17$

$2 = 3$

Subtract $15$ from $- 2$ .

$0 = 17$

$0 = - 17$

$0 = 17$

$2 = 3$

$0 = 17$

$0 = - 17$

$0 = 17$

$2 = 3$

Move all terms not containing a variable to the right side of the equation.

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Subtract $2$ from both sides of the equation.

$0 = 17$

$0 = - 17$

$0 = 3 - 2$

$0 = 17$

$0 = - 17$

Subtract $2$ from $3$ .

$0 = 17$

$0 = - 17$

$0 = 1$

$0 = 17$

$0 = - 17$

$0 = 17$

$0 = - 17$

$0 = 1$

$0 = 17$

$0 = - 17$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{17} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{17} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{1} = \frac{1}{17} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{17} R_{1}$ .

$[\begin{array}{c} (\frac{1}{17}) \cdot (17) \\ - 17 \\ 1 \end{array}]$

$R_{1} = \frac{1}{17} R_{1}$

Simplify $R_{1}$ (row $1$ ).

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

Perform the row operation $R_{2} = 17 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 17 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 17 \cdot R_{1} + R_{2} \\ 1 \end{array}]$

$R_{2} = 17 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 17 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 \\ (17) \cdot (1) - 17 \\ 1 \end{array}]$

$R_{2} = 17 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{3} = - 1 \cdot R_{1} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 1 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 1 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 1 \cdot R_{1} + R_{3}$ .

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

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

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

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

$0 = 1$

This expression is the solution set for the system of equations.

${}$

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.

$X = = [\begin{array}{c} 0 \end{array}]$

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