Linear Algebra Examples

x - 7 y = - 35

$x - 7 y = - 35$ ,

3 x - 4 y = - 5

$3 x - 4 y = - 5$

Step 1

Write the system of equations in matrix form.

[\begin{matrix} 1 & - 7 & - 35 3 & - 4 & - 5 \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 3 & - 4 & - 5 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 2.1.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{matrix} 1 & - 7 & - 35 3 - 3 \cdot 1 & - 4 - 3 \cdot - 7 & - 5 - 3 \cdot - 35 \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 3 - 3 \cdot 1 & - 4 - 3 \cdot - 7 & - 5 - 3 \cdot - 35 \end{array}]$

Step 2.1.2

Simplify

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & - 7 & - 35 0 & 17 & 100 \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & 17 & 100 \end{array}]$

[\begin{matrix} 1 & - 7 & - 35 0 & 17 & 100 \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & 17 & 100 \end{array}]$

Step 2.2

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{17}

$\frac{1}{17}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 2.2.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{17}

$\frac{1}{17}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{matrix} 1 & - 7 & - 35 \frac{0}{17} & \frac{17}{17} & \frac{100}{17} \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ \frac{0}{17} & \frac{17}{17} & \frac{100}{17} \end{array}]$

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & - 7 & - 35 0 & 1 & \frac{100}{17} \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & 1 & \frac{100}{17} \end{array}]$

[\begin{matrix} 1 & - 7 & - 35 0 & 1 & \frac{100}{17} \end{matrix}]

$[\begin{array}{c} 1 & - 7 & - 35 \\ 0 & 1 & \frac{100}{17} \end{array}]$

Step 2.3

Perform the row operation

R_{1} = R_{1} + 7 R_{2}

$R_{1} = R_{1} + 7 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 2.3.1

Perform the row operation

R_{1} = R_{1} + 7 R_{2}

$R_{1} = R_{1} + 7 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 + 7 \cdot 0 & - 7 + 7 \cdot 1 & - 35 + 7 (\frac{100}{17}) 0 & 1 & \frac{100}{17} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 + 7 \cdot 0 & - 7 + 7 \cdot 1 & - 35 + 7 (\frac{100}{17}) \\ 0 & 1 & \frac{100}{17} \end{array}]$

Step 2.3.2

Simplify

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & 0 & \frac{105}{17} 0 & 1 & \frac{100}{17} \end{matrix}]

$[\begin{array}{c} 1 & 0 & \frac{105}{17} \\ 0 & 1 & \frac{100}{17} \end{array}]$

Step 3

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

$x = \frac{105}{17}$

$y = \frac{100}{17}$

Step 4

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

$(\frac{105}{17}, \frac{100}{17})$

Step 5

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{105}{17} \\ \frac{100}{17} \end{array}]$

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