Linear Algebra Examples

$4 x - 3 y = 0$ , $- 4 x + 3 y = 0$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & - 3 & 0 \\ - 4 & 3 & 0 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

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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 $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 3}{4} & \frac{0}{4} \\ - 4 & 3 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & 0 \\ - 4 & 3 & 0 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{3}{4} & 0 \\ - 4 + 4 \cdot 1 & 3 + 4 (- \frac{3}{4}) & 0 + 4 \cdot 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{4} & 0 \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x - \frac{3 y}{4} = 0$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$- \frac{3 y}{4} = - x$

Step 4.2

Multiply both sides of the equation by $- \frac{4}{3}$ .

$- \frac{4}{3} (- \frac{3 y}{4}) = - \frac{4}{3} (- x)$

Step 4.3

Simplify both sides of the equation.

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Step 4.3.1

Simplify the left side.

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Step 4.3.1.1

Simplify $- \frac{4}{3} (- \frac{3 y}{4})$ .

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Step 4.3.1.1.1

Cancel the common factor of $4$ .

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Step 4.3.1.1.1.1

Move the leading negative in $- \frac{4}{3}$ into the numerator.

$\frac{- 4}{3} (- \frac{3 y}{4}) = - \frac{4}{3} (- x)$

Step 4.3.1.1.1.2

Move the leading negative in $- \frac{3 y}{4}$ into the numerator.

$\frac{- 4}{3} \cdot \frac{- 3 y}{4} = - \frac{4}{3} (- x)$

Step 4.3.1.1.1.3

Factor $4$ out of $- 4$ .

$\frac{4 (- 1)}{3} \cdot \frac{- 3 y}{4} = - \frac{4}{3} (- x)$

Step 4.3.1.1.1.4

Cancel the common factor.

$\frac{4 \cdot - 1}{3} \cdot \frac{- 3 y}{4} = - \frac{4}{3} (- x)$

Step 4.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{3} (- 3 y) = - \frac{4}{3} (- x)$

Step 4.3.1.1.2

Cancel the common factor of $3$ .

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Step 4.3.1.1.2.1

Factor $3$ out of $- 3 y$ .

$\frac{- 1}{3} (3 (- y)) = - \frac{4}{3} (- x)$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{3} (3 (- y)) = - \frac{4}{3} (- x)$

Step 4.3.1.1.2.3

Rewrite the expression.

$- - y = - \frac{4}{3} (- x)$

Step 4.3.1.1.3

Multiply.

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Step 4.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{4}{3} (- x)$

Step 4.3.1.1.3.2

Multiply $y$ by $1$ .

$y = - \frac{4}{3} (- x)$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Multiply $- \frac{4}{3} (- x)$ .

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Step 4.3.2.1.1

Multiply $- 1$ by $- 1$ .

$y = 1 (\frac{4}{3}) x$

Step 4.3.2.1.2

Multiply $\frac{4}{3}$ by $1$ .

$y = \frac{4}{3} x$

Step 4.3.2.1.3

Combine $\frac{4}{3}$ and $x$ .

$y = \frac{4 x}{3}$

Step 5

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

$(x, \frac{4 x}{3})$

Step 6

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ \frac{4 x}{3} \end{array}]$