प्री-कैलकुलस उदाहरण

$4 x - 8 y + 3 z = - 15$ , $x + 4 y - z = 0$ , $8 x - y - z = 3$

चरण 1

Write the system as a matrix.

$[\begin{array}{c} 4 & - 8 & 3 & - 15 \\ 1 & 4 & - 1 & 0 \\ 8 & - 1 & - 1 & 3 \end{array}]$

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 $\frac{1}{4}$ 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 $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 8}{4} & \frac{3}{4} & \frac{- 15}{4} \\ 1 & 4 & - 1 & 0 \\ 8 & - 1 & - 1 & 3 \end{array}]$

चरण 2.1.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 1 & 4 & - 1 & 0 \\ 8 & - 1 & - 1 & 3 \end{array}]$

चरण 2.2

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

और स्टेप्स के लिए टैप करें…

चरण 2.2.1

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

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 1 - 1 & 4 + 2 & - 1 - \frac{3}{4} & 0 + \frac{15}{4} \\ 8 & - 1 & - 1 & 3 \end{array}]$

चरण 2.2.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 6 & - \frac{7}{4} & \frac{15}{4} \\ 8 & - 1 & - 1 & 3 \end{array}]$

चरण 2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 6 & - \frac{7}{4} & \frac{15}{4} \\ 8 - 8 \cdot 1 & - 1 - 8 \cdot - 2 & - 1 - 8 (\frac{3}{4}) & 3 - 8 (- \frac{15}{4}) \end{array}]$

चरण 2.3.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 6 & - \frac{7}{4} & \frac{15}{4} \\ 0 & 15 & - 7 & 33 \end{array}]$

चरण 2.4

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ \frac{0}{6} & \frac{6}{6} & \frac{- \frac{7}{4}}{6} & \frac{\frac{15}{4}}{6} \\ 0 & 15 & - 7 & 33 \end{array}]$

चरण 2.4.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & - \frac{7}{24} & \frac{5}{8} \\ 0 & 15 & - 7 & 33 \end{array}]$

चरण 2.5

Perform the row operation $R_{3} = R_{3} - 15 R_{2}$ to make the entry at $3, 2$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.5.1

Perform the row operation $R_{3} = R_{3} - 15 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & - \frac{7}{24} & \frac{5}{8} \\ 0 - 15 \cdot 0 & 15 - 15 \cdot 1 & - 7 - 15 (- \frac{7}{24}) & 33 - 15 (\frac{5}{8}) \end{array}]$

चरण 2.5.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & - \frac{7}{24} & \frac{5}{8} \\ 0 & 0 & - \frac{21}{8} & \frac{189}{8} \end{array}]$

चरण 2.6

Multiply each element of $R_{3}$ by $- \frac{8}{21}$ to make the entry at $3, 3$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.6.1

Multiply each element of $R_{3}$ by $- \frac{8}{21}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & - \frac{7}{24} & \frac{5}{8} \\ - \frac{8}{21} \cdot 0 & - \frac{8}{21} \cdot 0 & - \frac{8}{21} (- \frac{21}{8}) & - \frac{8}{21} \cdot \frac{189}{8} \end{array}]$

चरण 2.6.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & - \frac{7}{24} & \frac{5}{8} \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.7

Perform the row operation $R_{2} = R_{2} + \frac{7}{24} R_{3}$ to make the entry at $2, 3$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{7}{24} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 + \frac{7}{24} \cdot 0 & 1 + \frac{7}{24} \cdot 0 & - \frac{7}{24} + \frac{7}{24} \cdot 1 & \frac{5}{8} + \frac{7}{24} \cdot - 9 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.7.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & \frac{3}{4} & - \frac{15}{4} \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.8

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

और स्टेप्स के लिए टैप करें…

चरण 2.8.1

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

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & - 2 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & - \frac{15}{4} - \frac{3}{4} \cdot - 9 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.8.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & - 2 & 0 & 3 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.9

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 2.9.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 0 + 2 \cdot 0 & 3 + 2 \cdot - 2 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 2.9.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & 0 & - 1 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - 9 \end{array}]$

चरण 3

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

$x = - 1$

$y = - 2$

$z = - 9$

चरण 4

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

$(- 1, - 2, - 9)$