Linear Algebra Examples

$x + y + 5 z = - 3$ , $x + 2 y - 2 z = 3$ , $4 x + 11 y + k z = 31$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 1 & 2 & - 2 & 3 \\ 4 & 11 & k & 31 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 1 - 1 & 2 - 1 & - 2 - 5 & 3 + 3 \\ 4 & 11 & k & 31 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 4 & 11 & k & 31 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 4 - 4 \cdot 1 & 11 - 4 \cdot 1 & k - 4 \cdot 5 & 31 - 4 \cdot - 3 \end{array}]$

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 0 & 7 & k - 20 & 43 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & k - 20 - 7 \cdot - 7 & 43 - 7 \cdot 6 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 0 & 0 & k + 29 & 1 \end{array}]$

Step 2.4

Multiply each element of $R_{3}$ by $\frac{1}{k + 29}$ to make the entry at $3, 3$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{3}$ by $\frac{1}{k + 29}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ \frac{0}{k + 29} & \frac{0}{k + 29} & \frac{k + 29}{k + 29} & \frac{1}{k + 29} \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & - 7 & 6 \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.5

Perform the row operation $R_{2} = R_{2} + 7 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{2} = R_{2} + 7 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 + 7 \cdot 0 & 1 + 7 \cdot 0 & - 7 + 7 \cdot 1 & 6 + 7 \frac{1}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 5 & - 3 \\ 0 & 1 & 0 & \frac{6 k + 181}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.6

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

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Step 2.6.1

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

$[\begin{array}{c} 1 - 5 \cdot 0 & 1 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 3 - 5 \frac{1}{k + 29} \\ 0 & 1 & 0 & \frac{6 k + 181}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & - \frac{3 k + 92}{k + 29} \\ 0 & 1 & 0 & \frac{6 k + 181}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.7

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

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Step 2.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & - \frac{3 k + 92}{k + 29} - \frac{6 k + 181}{k + 29} \\ 0 & 1 & 0 & \frac{6 k + 181}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 2.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3 (3 k + 91)}{k + 29} \\ 0 & 1 & 0 & \frac{6 k + 181}{k + 29} \\ 0 & 0 & 1 & \frac{1}{k + 29} \end{array}]$

Step 3

Since $\frac{1}{k + 29}$ is undefined when $k = - 29$ , then $k = - 29$ makes the system have no solution.

$k = - 29$

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