लीनियर एलजेब्रा उदाहरण

$s = [\begin{array}{c} 2 & 4 & 6 & 4 \\ 2 & 5 & 7 & 6 \\ 2 & 3 & 5 & 2 \end{array}]$

चरण 1

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

$[\begin{array}{c} 2 & 4 & 6 & 4 & 0 \\ 2 & 5 & 7 & 6 & 0 \\ 2 & 3 & 5 & 2 & 0 \end{array}]$

चरण 2

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 2.1

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

और स्टेप्स के लिए टैप करें…

चरण 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{4}{2} & \frac{6}{2} & \frac{4}{2} & \frac{0}{2} \\ 2 & 5 & 7 & 6 & 0 \\ 2 & 3 & 5 & 2 & 0 \end{array}]$

चरण 2.1.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 2 & 3 & 2 & 0 \\ 2 & 5 & 7 & 6 & 0 \\ 2 & 3 & 5 & 2 & 0 \end{array}]$

चरण 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & 2 & 3 & 2 & 0 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 2 & 7 - 2 \cdot 3 & 6 - 2 \cdot 2 & 0 - 2 \cdot 0 \\ 2 & 3 & 5 & 2 & 0 \end{array}]$

चरण 2.2.2

$R_{2}$ को सरल करें.

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

चरण 2.3

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

और स्टेप्स के लिए टैप करें…

चरण 2.3.1

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

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

चरण 2.3.2

$R_{3}$ को सरल करें.

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

चरण 2.4

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

और स्टेप्स के लिए टैप करें…

चरण 2.4.1

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

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

चरण 2.4.2

$R_{3}$ को सरल करें.

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

चरण 2.5

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

और स्टेप्स के लिए टैप करें…

चरण 2.5.1

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

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

चरण 2.5.2

$R_{1}$ को सरल करें.

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

चरण 3

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

$x_{1} + x_{3} - 2 x_{4} = 0$

$x_{2} + x_{3} + 2 x_{4} = 0$

$0 = 0$

चरण 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - x_{3} + 2 x_{4} \\ - x_{3} - 2 x_{4} \\ x_{3} \\ x_{4} \end{array}]$

चरण 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{3} [\begin{array}{c} - 1 \\ - 1 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 2 \\ - 2 \\ 0 \\ 1 \end{array}]$

चरण 6

Write as a solution set.

${x_{3} [\begin{array}{c} - 1 \\ - 1 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 2 \\ - 2 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$