फाइनाइट मैथ उदाहरण

x + 3 y - z = 4

$x + 3 y - z = 4$ ,

3 y - z = 0

$3 y - z = 0$ ,

x - y + 5 z = 0

$x - y + 5 z = 0$

चरण 1

Write the system as a matrix.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 3 & - 1 & 0 1 & - 1 & 5 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

चरण 2

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 2.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.1.1

Perform the row operation

R_{3} = R_{3} - R_{1}

$R_{3} = R_{3} - R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 3 & - 1 & 0 1 - 1 & - 1 - 3 & 5 + 1 & 0 - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 3 & - 1 & 0 \\ 1 - 1 & - 1 - 3 & 5 + 1 & 0 - 4 \end{array}]$

चरण 2.1.2

R_{3}

$R_{3}$ को सरल करें.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 3 & - 1 & 0 0 & - 4 & 6 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 3 & - 1 & 0 0 & - 4 & 6 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

चरण 2.2

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{3}

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

2, 2

$2, 2$ a

1

$1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.2.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{3}

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

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 \frac{0}{3} & \frac{3}{3} & \frac{- 1}{3} & \frac{0}{3} 0 & - 4 & 6 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

चरण 2.2.2

R_{2}

$R_{2}$ को सरल करें.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & - 4 & 6 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & - 4 & 6 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

चरण 2.3

Perform the row operation

R_{3} = R_{3} + 4 R_{2}

$R_{3} = R_{3} + 4 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.3.1

Perform the row operation

R_{3} = R_{3} + 4 R_{2}

$R_{3} = R_{3} + 4 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 6 + 4 (- \frac{1}{3}) & - 4 + 4 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 6 + 4 (- \frac{1}{3}) & - 4 + 4 \cdot 0 \end{array}]$

चरण 2.3.2

R_{3}

$R_{3}$ को सरल करें.

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & 0 & \frac{14}{3} & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & \frac{14}{3} & - 4 \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & 0 & \frac{14}{3} & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & \frac{14}{3} & - 4 \end{array}]$

चरण 2.4

Multiply each element of

R_{3}

$R_{3}$ by

\frac{3}{14}

$\frac{3}{14}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.4.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{3}{14}

$\frac{3}{14}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot \frac{14}{3} & \frac{3}{14} \cdot - 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot 0 & \frac{3}{14} \cdot \frac{14}{3} & \frac{3}{14} \cdot - 4 \end{array}]$

चरण 2.4.2

R_{3}

$R_{3}$ को सरल करें.

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & 0 & 1 & - \frac{6}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 & 1 & - \frac{1}{3} & 0 0 & 0 & 1 & - \frac{6}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & - \frac{1}{3} & 0 \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.5

Perform the row operation

R_{2} = R_{2} + \frac{1}{3} R_{3}

$R_{2} = R_{2} + \frac{1}{3} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.5.1

Perform the row operation

R_{2} = R_{2} + \frac{1}{3} R_{3}

$R_{2} = R_{2} + \frac{1}{3} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 3 & - 1 & 4 0 + \frac{1}{3} \cdot 0 & 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & 0 + \frac{1}{3} (- \frac{6}{7}) 0 & 0 & 1 & - \frac{6}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 + \frac{1}{3} \cdot 0 & 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & 0 + \frac{1}{3} (- \frac{6}{7}) \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.5.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & 3 & - 1 & 4 \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.6

Perform the row operation

$R_{1} = R_{1} + R_{3}$ to make the entry at

$1, 3$ a

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.6.1

Perform the row operation

$R_{1} = R_{1} + R_{3}$ to make the entry at

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 + 0 & 3 + 0 & - 1 + 1 \cdot 1 & 4 - \frac{6}{7} \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.6.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 3 & 0 & \frac{22}{7} \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.7

Perform the row operation

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.7.1

Perform the row operation

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 & \frac{22}{7} - 3 (- \frac{2}{7}) \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 2.7.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & - \frac{2}{7} \\ 0 & 0 & 1 & - \frac{6}{7} \end{array}]$

चरण 3

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

$x = 4$

$y = - \frac{2}{7}$

$z = - \frac{6}{7}$

चरण 4

The solution is the set of ordered pairs that make the system true.

$(4, - \frac{2}{7}, - \frac{6}{7})$

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