Linear Algebra Examples

$4 x + 2 = 5 y - 3$ , $y = 3 x - 1$

Step 1

Subtract $2$ from both sides of the equation.

$4 x = 5 y - 3 - 2, y = 3 x - 1$

Step 2

Subtract $5 y$ from both sides of the equation.

$4 x - 5 y = - 3 - 2, y = 3 x - 1$

Step 3

Subtract $3 x$ from both sides of the equation.

$4 x - 5 y = - 3 - 2, y - 3 x = - 1$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & - 5 & - 3 - 2 \\ - 3 & 1 & - 1 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Subtract $2$ from $- 3$ .

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

Step 5.2

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

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Step 5.2.1

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 5.2.2

Simplify $R_{1}$ .

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

Step 5.3

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 5.3.1

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 5.3.2

Simplify $R_{2}$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify $R_{2}$ .

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

Step 5.5

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 5.5.1

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 5.5.2

Simplify $R_{1}$ .

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

Step 6

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

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

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

Step 7

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

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

Step 8

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{10}{11} \\ \frac{19}{11} \end{array}]$