लीनियर एलजेब्रा उदाहरण

$s = {[\begin{array}{c} 6 \\ 2 \\ 1 \end{array}], [\begin{array}{c} 4 \\ - 3 \\ 1 \end{array}]}$

चरण 1

To determine if the columns in the matrix are linearly dependent, determine if the equation $A x = 0$ has any non-trivial solutions.

चरण 2

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 6 & 4 & 0 \\ 2 & - 3 & 0 \\ 1 & 1 & 0 \end{array}]$

चरण 3

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 3.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{4}{6} & \frac{0}{6} \\ 2 & - 3 & 0 \\ 1 & 1 & 0 \end{array}]$

चरण 3.1.2

$R_{1}$ को सरल करें.

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

चरण 3.2

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

और स्टेप्स के लिए टैप करें…

चरण 3.2.1

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

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

चरण 3.2.2

$R_{2}$ को सरल करें.

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

चरण 3.3

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

और स्टेप्स के लिए टैप करें…

चरण 3.3.1

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

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

चरण 3.3.2

$R_{3}$ को सरल करें.

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

चरण 3.4

Multiply each element of $R_{2}$ by $- \frac{3}{13}$ to make the entry at $2, 2$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 3.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{13}$ to make the entry at $2, 2$ a $1$ .

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

चरण 3.4.2

$R_{2}$ को सरल करें.

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

चरण 3.5

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

और स्टेप्स के लिए टैप करें…

चरण 3.5.1

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

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

चरण 3.5.2

$R_{3}$ को सरल करें.

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

चरण 3.6

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

और स्टेप्स के लिए टैप करें…

चरण 3.6.1

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

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

चरण 3.6.2

$R_{1}$ को सरल करें.

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

चरण 4

Remove rows that are all zeros.

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

चरण 5

Write the matrix as a system of linear equations.

$x = 0$

$y = 0$

चरण 6

Since the only solution to $A x = 0$ is the trivial solution, the vectors are linearly independent.

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