Linear Algebra Examples Systems of Linear Equations

Determine the Value of k for Which the System Has No Solutions

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

Step 1

Subtract $k$ from both sides of the equation.

$x + 2 y + z = 3, 2 x - y - 3 z = 5, 4 x + 3 y - z - k = 0$

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

Rewrite $- 1 k$ as $- k$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{2}$ .

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

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 & - 5 & - 5 & - 1 \\ - k + k \cdot 1 & 3 + k \cdot 2 & - 1 + k \cdot 1 & 0 + k \cdot 3 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

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

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot - 1 \\ 0 & 3 + 2 k & - 1 + k & 3 k \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 & 1 & 1 & \frac{1}{5} \\ 0 & 0 & - k - 4 & \frac{13 k - 3}{5} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $\frac{1}{- k - 4}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $\frac{1}{- k - 4}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 & 1 & 1 & \frac{1}{5} \\ \frac{0}{- k - 4} & \frac{0}{- k - 4} & \frac{- k - 4}{- k - 4} & \frac{\frac{13 k - 3}{5}}{- k - 4} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 & 1 & 1 & \frac{1}{5} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 - 0 & 1 - 0 & 1 - 1 & \frac{1}{5} + \frac{13 k - 3}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 1 & 3 \\ 0 & 1 & 0 & \frac{14 k + 1}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 - 0 & 2 - 0 & 1 - 1 & 3 + \frac{13 k - 3}{5 (k + 4)} \\ 0 & 1 & 0 & \frac{14 k + 1}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 0 & \frac{28 k + 57}{5 (k + 4)} \\ 0 & 1 & 0 & \frac{14 k + 1}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.9

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

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Step 3.9.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{28 k + 57}{5 (k + 4)} - 2 \frac{14 k + 1}{5 (k + 4)} \\ 0 & 1 & 0 & \frac{14 k + 1}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 3.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{11}{k + 4} \\ 0 & 1 & 0 & \frac{14 k + 1}{5 (k + 4)} \\ 0 & 0 & 1 & - \frac{13 k - 3}{5 (k + 4)} \end{array}]$

Step 4

Since $- \frac{13 k - 3}{5 (k + 4)}$ is undefined when $k = - 4$ , then $k = - 4$ makes the system have no solution.

$k = - 4$

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Determine the Value of k for Which the System Has No Solutions

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