लीनियर एलजेब्रा उदाहरण

$[\begin{array}{c} 6 & - 8 & 10 \\ - \frac{3}{4} & 1 & - \frac{5}{4} \end{array}]$

चरण 1

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

और स्टेप्स के लिए टैप करें…

चरण 1.1

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

$[\begin{array}{c} \frac{6}{6} & \frac{- 8}{6} & \frac{10}{6} \\ - \frac{3}{4} & 1 & - \frac{5}{4} \end{array}]$

चरण 1.2

$R_{1}$ को सरल करें.

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

चरण 2

Perform the row operation $R_{2} = R_{2} + \frac{3}{4} R_{1}$ to make the entry at $2, 1$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.1

Perform the row operation $R_{2} = R_{2} + \frac{3}{4} R_{1}$ to make the entry at $2, 1$ a $0$ .

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

चरण 2.2

$R_{2}$ को सरल करें.

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