Linear Algebra Examples

$4 x - 5 y = - 5$ , $3 x - y = 1$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify $R_{1}$ .

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

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{5}{4} \\ 3 - 3 \cdot 1 & - 1 - 3 (- \frac{5}{4}) & 1 - 3 (- \frac{5}{4}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{5}{4} \\ 0 & \frac{11}{4} & \frac{19}{4} \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{4}{11}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{4}{11}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{5}{4} \\ \frac{4}{11} \cdot 0 & \frac{4}{11} \cdot \frac{11}{4} & \frac{4}{11} \cdot \frac{19}{4} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{5}{4} \\ 0 & 1 & \frac{19}{11} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{4} \cdot 0 & - \frac{5}{4} + \frac{5}{4} \cdot 1 & - \frac{5}{4} + \frac{5}{4} \cdot \frac{19}{11} \\ 0 & 1 & \frac{19}{11} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{10}{11} \\ 0 & 1 & \frac{19}{11} \end{array}]$

Step 3

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

$x = \frac{10}{11}$

$y = \frac{19}{11}$

Step 4

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

$(\frac{10}{11}, \frac{19}{11})$

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