Linear Algebra Examples

$7 a = 4 - b$ , $7 a = 2 - b$

Step 1

Add $b$ to both sides of the equation.

$7 a + b = 4, 7 a = 2 - b$

Step 2

Add $b$ to both sides of the equation.

$7 a + b = 4, 7 a + b = 2$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 7 & 1 & 4 \\ 7 & 1 & 2 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{1}{7} & \frac{4}{7} \\ 7 & 1 & 2 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & \frac{4}{7} \\ 7 & 1 & 2 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{7} & \frac{4}{7} \\ 7 - 7 \cdot 1 & 1 - 7 (\frac{1}{7}) & 2 - 7 (\frac{4}{7}) \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & \frac{4}{7} \\ 0 & 0 & - 2 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{2}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{7} & \frac{4}{7} \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & \frac{4}{7} \\ 0 & 0 & 1 \end{array}]$

Step 4.4

Perform the row operation $R_{1} = R_{1} - \frac{4}{7} R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 4.4.1

Perform the row operation $R_{1} = R_{1} - \frac{4}{7} R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{4}{7} \cdot 0 & \frac{1}{7} - \frac{4}{7} \cdot 0 & \frac{4}{7} - \frac{4}{7} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 4.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & 0 \\ 0 & 0 & 1 \end{array}]$

Step 5

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

$a + \frac{b}{7} = 0$

$0 = 1$

Step 6

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

$a = - \frac{b}{7}$

$0 = 1$

Step 7

Since $0 \neq 1$ , there are no solutions.

No solution