फाइनाइट मैथ उदाहरण

$3 x - 2 y - 8 z = 1$ , $9 x - 6 y - 24 z = - 2$ , $x - y + z = 1$

चरण 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 2 & - 8 & 1 \\ 9 & - 6 & - 24 & - 2 \\ 1 & - 1 & 1 & 1 \end{array}]$

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 $\frac{1}{3}$ 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 $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{- 8}{3} & \frac{1}{3} \\ 9 & - 6 & - 24 & - 2 \\ 1 & - 1 & 1 & 1 \end{array}]$

चरण 2.1.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 9 & - 6 & - 24 & - 2 \\ 1 & - 1 & 1 & 1 \end{array}]$

चरण 2.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 9 - 9 \cdot 1 & - 6 - 9 (- \frac{2}{3}) & - 24 - 9 (- \frac{8}{3}) & - 2 - 9 (\frac{1}{3}) \\ 1 & - 1 & 1 & 1 \end{array}]$

चरण 2.2.2

$R_{2}$ को सरल करें.

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

चरण 2.3

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 2.3.1

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

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

चरण 2.3.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & 0 & 0 & - 5 \\ 0 & - \frac{1}{3} & \frac{11}{3} & \frac{2}{3} \end{array}]$

चरण 2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & - \frac{1}{3} & \frac{11}{3} & \frac{2}{3} \\ 0 & 0 & 0 & - 5 \end{array}]$

चरण 2.5

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

और स्टेप्स के लिए टैप करें…

चरण 2.5.1

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

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ - 3 \cdot 0 & - 3 (- \frac{1}{3}) & - 3 (\frac{11}{3}) & - 3 (\frac{2}{3}) \\ 0 & 0 & 0 & - 5 \end{array}]$

चरण 2.5.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & 1 & - 11 & - 2 \\ 0 & 0 & 0 & - 5 \end{array}]$

चरण 2.6

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 4$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{5}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & 1 & - 11 & - 2 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 \end{array}]$

चरण 2.6.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & 1 & - 11 & - 2 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.7

Perform the row operation $R_{2} = R_{2} + 2 R_{3}$ to make the entry at $2, 4$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.7.1

Perform the row operation $R_{2} = R_{2} + 2 R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & - 11 + 2 \cdot 0 & - 2 + 2 \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.7.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & \frac{1}{3} \\ 0 & 1 & - 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.8

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

और स्टेप्स के लिए टैप करें…

चरण 2.8.1

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

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & - \frac{2}{3} - \frac{1}{3} \cdot 0 & - \frac{8}{3} - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 \\ 0 & 1 & - 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.8.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{8}{3} & 0 \\ 0 & 1 & - 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.9

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 2.9.1

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

$[\begin{array}{c} 1 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & - \frac{8}{3} + \frac{2}{3} \cdot - 11 & 0 + \frac{2}{3} \cdot 0 \\ 0 & 1 & - 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 2.9.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & - 10 & 0 \\ 0 & 1 & - 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 3

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

$x - 10 z = 0$

$y - 11 z = 0$

$0 = 1$

चरण 4

The system is inconsistent so there is no solution.

$कोई हल नहीं$