एलजेब्रा उदाहरण

$[\begin{array}{c} 5 & 11 & 3 \end{array}] - [\begin{array}{c} 2 & 0 & 2 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

चरण 1

संबंधित तत्वों को घटाएं.

$[\begin{array}{c} 5 - 2 & 11 - 0 & 3 - 2 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

चरण 2

Simplify each element.

और स्टेप्स के लिए टैप करें…

चरण 2.1

$5$ में से $2$ घटाएं.

$[\begin{array}{c} 3 & 11 - 0 & 3 - 2 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

चरण 2.2

$11$ में से $0$ घटाएं.

$[\begin{array}{c} 3 & 11 & 3 - 2 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

चरण 2.3

$3$ में से $2$ घटाएं.

$[\begin{array}{c} 3 & 11 & 1 \end{array}] = [\begin{array}{c} 3 & x & 1 \end{array}]$

चरण 3

Write as a linear system of equations.

$3 = 3$

$11 = x$

$1 = 1$

चरण 4

Since $1 = 1$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying $11 = x$ .