लीनियर एलजेब्रा उदाहरण

$[\begin{array}{c} 1 & h & 4 \\ 3 & 6 & 8 \end{array}]$

चरण 1

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

और स्टेप्स के लिए टैप करें…

चरण 1.1

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

$[\begin{array}{c} 1 & h & 4 \\ 3 - 3 \cdot 1 & 6 - 3 h & 8 - 3 \cdot 4 \end{array}]$

चरण 1.2

$R_{2}$ को सरल करें.

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

चरण 2

Multiply each element of $R_{2}$ by $\frac{1}{6 - 3 h}$ to make the entry at $2, 2$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.1

Multiply each element of $R_{2}$ by $\frac{1}{6 - 3 h}$ to make the entry at $2, 2$ a $1$ .

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

चरण 2.2

$R_{2}$ को सरल करें.

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

चरण 3

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

और स्टेप्स के लिए टैप करें…

चरण 3.1

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

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

चरण 3.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & 4 + \frac{4 h}{3 (2 - h)} \\ 0 & 1 & - \frac{4}{3 (2 - h)} \end{array}]$