Álgebra Ejemplos

$- 2 [\begin{array}{c} x & - 1 \\ 3 & 5 \end{array}] + [\begin{array}{c} 3 & 8 \\ - 1 & 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 1

Simplifica cada término.

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Paso 1.1

Multiplica $- 2$ por cada elemento de la matriz.

$[\begin{array}{c} - 2 x & - 2 \cdot - 1 \\ - 2 \cdot 3 & - 2 \cdot 5 \end{array}] + [\begin{array}{c} 3 & 8 \\ - 1 & 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 1.2

Simplifica cada elemento de la matriz.

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Paso 1.2.1

Multiplica $- 2$ por $- 1$ .

$[\begin{array}{c} - 2 x & 2 \\ - 2 \cdot 3 & - 2 \cdot 5 \end{array}] + [\begin{array}{c} 3 & 8 \\ - 1 & 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 1.2.2

Multiplica $- 2$ por $3$ .

$[\begin{array}{c} - 2 x & 2 \\ - 6 & - 2 \cdot 5 \end{array}] + [\begin{array}{c} 3 & 8 \\ - 1 & 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 1.2.3

Multiplica $- 2$ por $5$ .

$[\begin{array}{c} - 2 x & 2 \\ - 6 & - 10 \end{array}] + [\begin{array}{c} 3 & 8 \\ - 1 & 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 2

Suma los elementos correspondientes.

$[\begin{array}{c} - 2 x + 3 & 2 + 8 \\ - 6 - 1 & - 10 + 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 3

Simplify each element.

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Paso 3.1

Suma $2$ y $8$ .

$[\begin{array}{c} - 2 x + 3 & 10 \\ - 6 - 1 & - 10 + 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 3.2

Resta $1$ de $- 6$ .

$[\begin{array}{c} - 2 x + 3 & 10 \\ - 7 & - 10 + 6 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 3.3

Suma $- 10$ y $6$ .

$[\begin{array}{c} - 2 x + 3 & 10 \\ - 7 & - 4 \end{array}] = [\begin{array}{c} x & 10 \\ - 7 & - 4 \end{array}]$

Paso 4

Write as a linear system of equations.

$- 2 x + 3 = x$

$10 = 10$

$- 7 = - 7$

$- 4 = - 4$

Paso 5

Since $- 4 = - 4$ is always true, the matrix equation has infinite solutions.

Infinite solutions satisfying $- 2 x + 3 = x$ .