Matemática discreta Ejemplos

$[\begin{array}{c} 2 & 4 \\ - 6 & - 4 \\ x & 4 \end{array}] + [\begin{array}{c} 4 & 10 \\ - 4 & - 2 \\ - 5 & y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 1

Suma los elementos correspondientes.

$[\begin{array}{c} 2 + 4 & 4 + 10 \\ - 6 - 4 & - 4 - 2 \\ x - 5 & 4 + y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 2

Simplify each element.

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Paso 2.1

Suma $2$ y $4$ .

$[\begin{array}{c} 6 & 4 + 10 \\ - 6 - 4 & - 4 - 2 \\ x - 5 & 4 + y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 2.2

Suma $4$ y $10$ .

$[\begin{array}{c} 6 & 14 \\ - 6 - 4 & - 4 - 2 \\ x - 5 & 4 + y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 2.3

Resta $4$ de $- 6$ .

$[\begin{array}{c} 6 & 14 \\ - 10 & - 4 - 2 \\ x - 5 & 4 + y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 2.4

Resta $2$ de $- 4$ .

$[\begin{array}{c} 6 & 14 \\ - 10 & - 6 \\ x - 5 & 4 + y \end{array}] = [\begin{array}{c} 6 & 6 \\ - 10 & - 6 \\ - 4 & 2 \end{array}]$

Paso 3

Write as a linear system of equations.

$6 = 6$

$14 = 6$

$- 10 = - 10$

$- 6 = - 6$

$x - 5 = - 4$

$4 + y = 2$

Paso 4

Since $14 \neq 6$ , the matrix equation has no solution.

No hay solución