लीनियर एलजेब्रा उदाहरण

$5 a + 3 b + d = 90$ , $2 a + 4 b + g = 64$

चरण 1

Write the system as a matrix.

$[\begin{array}{c} 5 & 3 & 1 & 0 & 90 \\ 2 & 4 & 0 & 1 & 64 \end{array}]$

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 $\frac{1}{5}$ 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 $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{3}{5} & \frac{1}{5} & \frac{0}{5} & \frac{90}{5} \\ 2 & 4 & 0 & 1 & 64 \end{array}]$

चरण 2.1.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & 0 & 18 \\ 2 & 4 & 0 & 1 & 64 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & 0 & 18 \\ 2 - 2 \cdot 1 & 4 - 2 (\frac{3}{5}) & 0 - 2 (\frac{1}{5}) & 1 - 2 \cdot 0 & 64 - 2 \cdot 18 \end{array}]$

चरण 2.2.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & 0 & 18 \\ 0 & \frac{14}{5} & - \frac{2}{5} & 1 & 28 \end{array}]$

चरण 2.3

Multiply each element of $R_{2}$ by $\frac{5}{14}$ to make the entry at $2, 2$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.3.1

Multiply each element of $R_{2}$ by $\frac{5}{14}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & 0 & 18 \\ \frac{5}{14} \cdot 0 & \frac{5}{14} \cdot \frac{14}{5} & \frac{5}{14} (- \frac{2}{5}) & \frac{5}{14} \cdot 1 & \frac{5}{14} \cdot 28 \end{array}]$

चरण 2.3.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & \frac{3}{5} & \frac{1}{5} & 0 & 18 \\ 0 & 1 & - \frac{1}{7} & \frac{5}{14} & 10 \end{array}]$

चरण 2.4

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{5} \cdot 0 & \frac{3}{5} - \frac{3}{5} \cdot 1 & \frac{1}{5} - \frac{3}{5} (- \frac{1}{7}) & 0 - \frac{3}{5} \cdot \frac{5}{14} & 18 - \frac{3}{5} \cdot 10 \\ 0 & 1 & - \frac{1}{7} & \frac{5}{14} & 10 \end{array}]$

चरण 2.4.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & \frac{2}{7} & - \frac{3}{14} & 12 \\ 0 & 1 & - \frac{1}{7} & \frac{5}{14} & 10 \end{array}]$

चरण 3

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

$a + \frac{2}{7} d - \frac{3}{14} g = 12$

$b - \frac{1}{7} d + \frac{5}{14} g = 10$

चरण 4

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

$(12 - \frac{2 d}{7} + \frac{3 g}{14}, 10 + \frac{d}{7} - \frac{5 g}{14}, d, g)$