Linear Algebra Examples

$4 a x - 3 y = 7$ , $- x - b y = 8$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 x & 0 & 0 & - 3 & 7 \\ 0 & - 1 y & - 1 & 0 & 8 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Rewrite $- 1 y$ as $- y$ .

$[\begin{array}{c} 4 x & 0 & 0 & - 3 & 7 \\ 0 & - y & - 1 & 0 & 8 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} \frac{4 x}{4 x} & \frac{0}{4 x} & \frac{0}{4 x} & \frac{- 3}{4 x} & \frac{7}{4 x} \\ 0 & - y & - 1 & 0 & 8 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3}{4 x} & \frac{7}{4 x} \\ 0 & - y & - 1 & 0 & 8 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3}{4 x} & \frac{7}{4 x} \\ - \frac{1}{y} \cdot 0 & - \frac{1}{y} (- y) & - \frac{1}{y} \cdot - 1 & - \frac{1}{y} \cdot 0 & - \frac{1}{y} \cdot 8 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3}{4 x} & \frac{7}{4 x} \\ 0 & 1 & \frac{1}{y} & 0 & - \frac{8}{y} \end{array}]$

Step 3

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

$a - \frac{3 y}{4 x} = \frac{7}{4 x}$

$b + \frac{x}{y} = - \frac{8}{y}$

Step 4

Solve the equation for $a$ .

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Step 4.1

Subtract $\frac{7}{4 x}$ from both sides of the equation.

$a - \frac{3 y}{4 x} - \frac{7}{4 x} = 0$

$b + \frac{x}{y} = - \frac{8}{y}$

Step 4.2

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

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Step 4.2.1

Add $\frac{3 y}{4 x}$ to both sides of the equation.

$a - \frac{7}{4 x} = \frac{3 y}{4 x}$

$b + \frac{x}{y} = - \frac{8}{y}$

Step 4.2.2

Add $\frac{7}{4 x}$ to both sides of the equation.

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

$b + \frac{x}{y} = - \frac{8}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

$b + \frac{x}{y} = - \frac{8}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

$b + \frac{x}{y} = - \frac{8}{y}$

Step 5

Solve the equation for $b$ .

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Step 5.1

Add $\frac{8}{y}$ to both sides of the equation.

$b + \frac{x}{y} + \frac{8}{y} = 0$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

Step 5.2

Move all terms not containing $b$ to the right side of the equation.

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Step 5.2.1

Subtract $\frac{x}{y}$ from both sides of the equation.

$b + \frac{8}{y} = - \frac{x}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

Step 5.2.2

Subtract $\frac{8}{y}$ from both sides of the equation.

$b = - \frac{x}{y} - \frac{8}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

$b = - \frac{x}{y} - \frac{8}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

$b = - \frac{x}{y} - \frac{8}{y}$

$a = \frac{3 y}{4 x} + \frac{7}{4 x}$

Step 6

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

$(\frac{3 y}{4 x} + \frac{7}{4 x}, - \frac{x}{y} - \frac{8}{y}, x, y)$

Step 7

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.

$X = [\begin{array}{c} a \\ b \\ x \\ y \end{array}] = [\begin{array}{c} \frac{3 y}{4 x} + \frac{7}{4 x} \\ - \frac{x}{y} - \frac{8}{y} \\ x \\ y \end{array}]$