Linear Algebra Examples

$x ([\begin{array}{c} a \\ b \\ d \end{array}]) = [\begin{array}{c} 5 a + 3 b - 2 d \\ 0 a + 4 b - d \end{array}]$ , $[\begin{array}{c} 14 \\ 2 \end{array}]$

Step 1

Write as an augmented matrix for $x ([\begin{array}{c} a \\ b \\ d \end{array}]) x = [\begin{array}{c} 14 \\ 2 \end{array}]$ .

$[\begin{array}{c} 5 a + 3 b - 2 d & 14 \\ 0 a + 4 b - d & 2 \end{array}]$

Step 2

Write as a linear system of equations.

$5 a + 3 b - 2 d = 14$

$0 a + 4 b - d = 2$

Step 3

Solve the system of equations.

Tap for more steps...

Step 3.1

Move variables to the left and constant terms to the right.

Tap for more steps...

Step 3.1.1

Reorder factors in $0 a + 4 b - d = 2$ .

$5 a + 3 b - 2 d = 14$

$a \cdot 0 + 4 b - d = 2$

Step 3.1.2

Move $a \cdot 0$ .

$5 a + 3 b - 2 d = 14$

$4 b - d + a \cdot 0 = 2$

$5 a + 3 b - 2 d = 14$

$4 b - d + a \cdot 0 = 2$

Step 3.2

Write the system as a matrix.

$[\begin{array}{c} 5 & 3 & - 2 & 14 \\ 4 & - 1 & 0 & 2 \end{array}]$

Step 3.3

Find the reduced row echelon form.

Tap for more steps...

Step 3.3.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 3.3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{3}{5} & \frac{- 2}{5} & \frac{14}{5} \\ 4 & - 1 & 0 & 2 \end{array}]$

Step 3.3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 4 & - 1 & 0 & 2 \end{array}]$

Step 3.3.2

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

Tap for more steps...

Step 3.3.2.1

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

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 4 - 4 \cdot 1 & - 1 - 4 (\frac{3}{5}) & 0 - 4 (- \frac{2}{5}) & 2 - 4 (\frac{14}{5}) \end{array}]$

Step 3.3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 0 & - \frac{17}{5} & \frac{8}{5} & - \frac{46}{5} \end{array}]$

Step 3.3.3

Multiply each element of $R_{2}$ by $- \frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 3.3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ - \frac{5}{17} \cdot 0 & - \frac{5}{17} (- \frac{17}{5}) & - \frac{5}{17} \cdot \frac{8}{5} & - \frac{5}{17} (- \frac{46}{5}) \end{array}]$

Step 3.3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{5} & - \frac{2}{5} & \frac{14}{5} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Step 3.3.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 3.3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & - \frac{2}{5} - \frac{3}{5} (- \frac{8}{17}) & \frac{14}{5} - \frac{3}{5} \cdot \frac{46}{17} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Step 3.3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{2}{17} & \frac{20}{17} \\ 0 & 1 & - \frac{8}{17} & \frac{46}{17} \end{array}]$

Step 3.4

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

$a - \frac{2}{17} d = \frac{20}{17}$

$b - \frac{8}{17} d = \frac{46}{17}$

Step 3.5

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} a \\ b \\ d \end{array}] = [\begin{array}{c} \frac{20}{17} + \frac{2 d}{17} \\ \frac{46}{17} + \frac{8 d}{17} \\ d \end{array}]$

Step 3.6

Write the solution as a linear combination of vectors.

$[\begin{array}{c} a \\ b \\ d \end{array}] = [\begin{array}{c} \frac{20}{17} \\ \frac{46}{17} \\ 0 \end{array}] + d [\begin{array}{c} \frac{2}{17} \\ \frac{8}{17} \\ 1 \end{array}]$

Step 3.7

Write as a solution set.

${[\begin{array}{c} \frac{20}{17} \\ \frac{46}{17} \\ 0 \end{array}] + d [\begin{array}{c} \frac{2}{17} \\ \frac{8}{17} \\ 1 \end{array}] | d \in R}$